【数学I】1章：数と式（節末問題・章末問題）

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% 節末問題I1.1.1：（One More）★
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ある多項式に$5x^2-3x+1$を加えるところを誤って引いたので，答えが$-3x^2+12x-5$になった．正しい答えを求めよ．

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% 節末問題I1.1.1の解答（One More）★
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多項式を$P$とおくと，$P-(5x^2-3x+1)=-3x^2+12x-5$

したがって，$P=-3x^2+12x-5+(5x^2-3x+1)=2x^2+9x-4$

よって，正しい答えは，

\begin{align*}
P+(5x^2-3x+1)&=2x^2+9x-4+5x^2-3x+1\\
&=7x^2+6x-3
\end{align*}

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% 節末問題I1.1.2：（One More）★★
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$(x^3-4x^2+2x+3)(x^3+x^2-x+2)$の展開式において，$x^5$と$x^3$の係数を求めよ．

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% 節末問題I1.1.2の解答（One More）★★
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$x^5$の項を計算すると，$x^3\cdot x^2+(-4x^2)\cdot x^3=(1-4)x^5=-3x^5$

よって，$x^5$の係数は$-3$

$x^3$の項を計算すると，

\[
x^3\cdot 2+(-4x^2)\cdot(-x)+(2x)\cdot x^2+3\cdot x^3=11x^3
\]

よって，$x^3$の係数は$11$

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% 節末問題I1.1.3：（One More）★★
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次の式を展開せよ．

(1) $(x^3+x^2+x+1)(x^3-x^2+x-1)$

(2) $(a+b+c)(b+c-a)(c+a-b)(a+b-c)$

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% 節末問題I1.1.3の解答（One More）★★
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(1)

\begin{align*}
&(x^3+x^2+x+1)(x^3-x^2+x-1)\\
=&\{(x^3+x)+(x^2+1)\}\{(x^3+x)-(x^2+1)\}\\
=&(x^3+x)^2-(x^2+1)^2\\
=&(x^6+2x^4+x^2)-(x^4+2x^2+1)\\
=&x^6+2x^4+x^2-x^4-2x^2-1\\
=&x^6+x^4-x^2-1
\end{align*}

(2)

\begin{align*}
&(a+b+c)(b+c-a)(c+a-b)(a+b-c)\\
=&\{(b+c)+a\}\{(b+c)-a\}\times\{a-(b-c)\}\{a+(b-c)\}\\
=&\{(b+c)^2-a^2\}\{a^2-(b-c)^2\}\\
=&-a^4+\{(b+c)^2+(b-c)^2\}a^2-(b+c)^2(b-c)^2\\
=&-a^4+2(b^2+c^2)a^2-(b^2-c^2)^2\\
=&-a^4+2a^2b^2+2c^2a^2-b^4+2b^2c^2-c^4\\
=&-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2
\end{align*}

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% 節末問題I1.1.4：（One More）★★
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次の式を因数分解せよ．

(1) $x^2y+2xy^2+x^2+4y^2+3xy+x+2y-2$

(2) $(x+y)^4-(x-y)^4$

(3) $(x+y)^3+z^3$

(4) $x^6-1$

(5) $a^6-7a^3-8$

(6) $(x^2+6x+3)(x^2+6x+7)+4$

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% 節末問題I1.1.4の解答（One More）★★
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(1)

\begin{align*}
&x^2y+2xy^2+x^2+4y^2+3xy+x+2y-2\\
=&(y+1)x^2+(2y^2+3y+1)x+4y^2+2y-2\\
=&(y+1)x^2+(y+1)(2y+1)x+2(y+1)(2y-1)\\
=&(y+1)\{x^2+(2y+1)x+2(2y-1)\}\\
=&(y+1)(x+2)\{x+(2y-1)\}\\
=&(x+2)(y+1)(x+2y-1)
\end{align*}

(2)

\begin{align*}
&(x+y)^4-(x-y)^4\\
=&\{(x+y)^2+(x-y)^2\}\{(x+y)^2-(x-y)^2\}\\
=&(x^2+2xy+y^2+x^2-2xy+y^2)\\
&\times\{(x+y)+(x-y)\}\{(x+y)-(x-y)\}\\
=&(2x^2+2y^2)(2x)(2y)=8xy(x^2+y^2)
\end{align*}

(3)

\begin{align*}
(x+y)^3+z^3&=\{(x+y)+z\}\{(x+y)^2-(x+y)\cdot z+z^2\}\\
&=(x+y+z)(x^2+2xy+y^2-xz-yz+z^2)
\end{align*}

(4)

\begin{align*}
x^6-1&=(x^3)^2-1^2\\
&=(x^3-1)(x^3+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1)
\end{align*}

(5)

\begin{align*}
&a^6-7a^3-8\\
=&(a^3+1)(a^3-8)\\
=&(a+1)(a^2-a+1)(a-2)(a^2+2a+4)\\
=&(a+1)(a-2)(a^2-a+1)(a^2+2a+4)
\end{align*}

(6)

\begin{align*}
&(x^2+6x+3)(x^2+6x+7)+4\\
=&\{(x^2+6x)+3\}\{(x^2+6x)+7\}+4\\
=&(x^2+6x)^2+10(x^2+6x)+25=(x^2+6x+5)^2\\
=&\{(x+1)(x+5)\}^2=(x+1)^2(x+5)^2
\end{align*}

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% 節末問題I1.1.5：（One More）★★★
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次の式を因数分解せよ．

(1) $(x-z)^3+(y-z)^3-(x+y-2z)^3$

(2) $4x^4+7x^2y^2+16y^4$

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(1)

\begin{align*}
&(x-z)^3+(y-z)^3-(x+y-2z)^3\\
&=(x-z)^3+(y-z)^3+(-x-y+2z)^3
\end{align*}

$x-z=a,y-z=b,-x-y+2z=c$とおくと，$a+b+c=0$より，

\begin{align*}
&(x-z)^3+(y-z)^3+(-x-y+2z)^3=a^3+b^3+c^3\\
&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc\\
&=3abc\\
&=3(x-z)(y-z)(-x-y+2z)\\
&=3(y-z)(z-x)(x+y-2z)
\end{align*}

(2)

\begin{align*}
4x^4+7x^2y^2+16y^4&=(4x^4+16x^2y^2+16y^4)-9x^2y^2\\
&=4(x^4+4x^2y^2+4y^4)-(3xy)^2\\
&=4(x^2+2y^2)^2-(3xy)^2\\
&=\{2(x^2+2y^2)+3xy\}\{2(x^2+2y^2)-3xy\}\\
&=(2x^2+3xy+4y^2)(2x^2-3xy+4y^2)
\end{align*}

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% 節末問題I1.2.1：（One More）★
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循環小数の積$0.\dot{1}\dot{5}\times 0.\dot{5}\dot{4}$を，1つの既約分数で表せ．

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% 節末問題I1.2.1の解答（One More）★
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$x=0.\dot{1}\dot{5}$とおくと，$100x=15.1515\ldots$

したがって，$100x-x=15$

これより，$x=\frac{15}{99}=\frac{5}{33}$

また，$y=0.\dot{5}\dot{4}$とおくと$100y=54.5454\ldots$

したがって，$100y-y=54$

これより，$y=\frac{54}{99}=\frac{6}{11}$

よって，

\[
0.\dot{1}\dot{5}\times 0.\dot{5}\dot{4}=xy=\frac{5}{33}\cdot\frac{6}{11}=\frac{30}{363}=\frac{10}{121}
\]

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% 節末問題I1.2.2：（One More）★★
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$\frac{3}{4}<x<\frac{5}{6}$のとき，$\sqrt{16x^2-24x+9}-\sqrt{x^2+6x+9}+\sqrt{36x^2-60x+25}$を簡単にせよ．

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% 節末問題I1.2.2の解答（One More）★★
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\begin{align*}
&\sqrt{16x^2-24x+9}-\sqrt{x^2+6x+9}+\sqrt{36x^2-60x+25}\\
=&\sqrt{(4x-3)^2}-\sqrt{(x+3)^2}+\sqrt{(6x-5)^2}
\end{align*}

$\frac{3}{4}<x<\frac{5}{6}$のとき，$4x-3>0,x+3>0,6x-5<0$であるから，

\begin{align*}
&\sqrt{16x^2-24x+9}-\sqrt{x^2+6x+9}+\sqrt{36x^2-60x+25}\\
=&(4x-3)-(x+3)+\{-(6x-5)\}\\
=&-3x-1
\end{align*}

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% 節末問題I1.2.3：（One More）★★
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次の式を計算せよ．

\[
S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}
\]

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% 節末問題I1.2.3の解答（One More）★★
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与えられた式の各項を有理化すると，

\[
S=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{100}-\sqrt{99})
\]

となり，途中の項が打ち消し合うから，

\[
S=-\sqrt{1}+\sqrt{100}
\]

よって，

\[
S=\sqrt{100}-\sqrt{1}=10-1=9
\]

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% 節末問題I1.2.4：（One More）★★
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次の式の分母を有理化して計算せよ．

\[
\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}
\]

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% 節末問題I1.2.4の解答（One More）★★
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\begin{align*}
&\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}\\
=&\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\{(\sqrt{2}+\sqrt{3})+\sqrt{5}\}\{(\sqrt{2}+\sqrt{3})-\sqrt{5}\}}\\
&+\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\{(\sqrt{2}-\sqrt{3})-\sqrt{5}\}\{(\sqrt{2}-\sqrt{3})+\sqrt{5}\}}\\
=&\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}+\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{(\sqrt{2}-\sqrt{3})^2-(\sqrt{5})^2}\\
=&\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\\
=&\frac{2\sqrt{3}-2\sqrt{5}}{2\sqrt{6}}=\frac{\sqrt{3}-\sqrt{5}}{\sqrt{6}}=\frac{(\sqrt{3}-\sqrt{5})\sqrt{6}}{6}=\frac{3\sqrt{2}-\sqrt{30}}{6}
\end{align*}

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% 節末問題I1.2.4の別解
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\begin{align*}
&\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}=\frac{(\sqrt{2}-\sqrt{3}-\sqrt{5})+(\sqrt{2}+\sqrt{3}+\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}-\sqrt{3}-\sqrt{5})}\\
=&\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{2}+\sqrt{3}+\sqrt{5}}{(\sqrt{2})^2-(\sqrt{3}+\sqrt{5})^2}=\frac{2\sqrt{2}}{-6-2\sqrt{15}}\\
=&-\frac{\sqrt{2}}{\sqrt{15}+3}=-\frac{\sqrt{2}(\sqrt{15}-3)}{(\sqrt{15}+3)(\sqrt{15}-3)}=\frac{3\sqrt{2}-\sqrt{30}}{6}
\end{align*}

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% 節末問題I1.2.5：（One More）★★
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次の式を簡単な形にせよ．

\[
\sqrt{4+4\sqrt{4+2\sqrt{3}}}
\]

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% 節末問題I1.2.5の解答（One More）★★
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\[
\sqrt{4+2\sqrt{3}}=\sqrt{(3+1)+2\sqrt{3\times 1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1
\]

よって，

\begin{align*}
\sqrt{4+4\sqrt{4+2\sqrt{3}}}&=\sqrt{4+4(\sqrt{3}+1)}\\
&=\sqrt{4+4\sqrt{3}+4}\\
&=\sqrt{8+4\sqrt{3}}\\
&=\sqrt{(6+2)+2\sqrt{6\times 2}}\\
&=\sqrt{(\sqrt{6}+\sqrt{2})^2}\\
&=\sqrt{6}+\sqrt{2}
\end{align*}

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% 節末問題I1.2.6：（One More）★★★
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実数$a,b,c$が$a+b+c=3,a^2+b^2+c^2=14,abc=-2$を満たすとき，$(a+b)(b+c)(c+a)$の値を求めよ．

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% 節末問題I1.2.6の解答（One More）★★★
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$a+b+c=3\cdots(\mathrm{i})$より，

\begin{align*}
(a+b)(b+c)(c+a)&=(3-c)(3-a)(3-b)\\
&=27-9a-9b-9c+3ab+3bc+3ca-abc\\
&=27-9(a+b+c)+3(ab+bc+ca)-abc\cdots(\mathrm{ii})
\end{align*}

また，$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$より，

\begin{align*}
ab+bc+ca&=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}\\
&=\frac{3^2-14}{2}=\frac{9-14}{2}=-\frac{5}{2}\cdots(\mathrm{iii})
\end{align*}

よって，$abc=-2$，(i)，(iii)を(ii)に代入すると，

\[
(a+b)(b+c)(c+a)=27-9\cdot 3+3\cdot(-\frac{5}{2})-(-2)=-\frac{11}{2}
\]

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% 節末問題I1.2.7：（One More）★★★
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$\frac{1}{4-\sqrt{11}}$の整数部分を$a$，小数部分を$b$とし，$10b$の整数部分を$c$，小数部分を$d$とするとき，次の値を求めよ．

(1) $a$

(2) $10b$

(3) $c$

(4) $d$

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(1)

\[
\frac{1}{4-\sqrt{11}}=\frac{4+\sqrt{11}}{(4-\sqrt{11})(4+\sqrt{11})}=\frac{4+\sqrt{11}}{5}
\]

$3<\sqrt{11}<4$であるから，$7<4+\sqrt{11}<8$

したがって，$\frac{7}{5}<\frac{4+\sqrt{11}}{5}<\frac{8}{5}$

よって，$a=1$

(2) $a+b=\frac{4+\sqrt{11}}{5}$であるから，

\[
b=\frac{4+\sqrt{11}}{5}-a=\frac{4+\sqrt{11}}{5}-1=\frac{-1+\sqrt{11}}{5}
\]

よって，$10b=10\cdot\frac{-1+\sqrt{11}}{5}=2(-1+\sqrt{11})=-2+2\sqrt{11}$

(3) (2)より，$10b=-2+2\sqrt{11}=2\sqrt{11}-2$

ここで，$6<2\sqrt{11}<7$であるから，$4<2\sqrt{11}-2<5$

よって，$c=4$

(4) $10b=c+d$より，$c+d=2\sqrt{11}-2$

よって，$d=(2\sqrt{11}-2)-c=(2\sqrt{11}-2)-4=2\sqrt{11}-6$

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% 節末問題I1.3.1：（One More）★★
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連立不等式$\{\begin{array}{l}x>4a-3\\
3x-2>8(x-1)\end{array}$の解について，次の条件を満たす定数$a$の値の範囲を求めよ．

(1) 解に0が含まれる．

(2) 解に含まれる整数がちょうど4個存在する．

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% 節末問題I1.3.1の解答（One More）★★
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$x>4a-3\cdots(\mathrm{i})$とする．

$3x-2>8(x-1)$より，$3x-2>8x-8$

よって，$x<\frac{6}{5}\cdots(\mathrm{ii})$

(1) $x=0$は(ii)に含まれるから，$x=0$が(i)の解に含まれる範囲を考える．

このとき，$4a-3<0$

よって，$a<\frac{3}{4}$

(2) (i)，(ii)を同時に満たす整数が存在するから，(i)と(ii)に共通範囲があり，

\[
4a-3<x<\frac{6}{5}
\]

$\frac{6}{5}=1.2$であるから，(i)，(ii)より，不等式を満たす整数がちょうど4個となるのは右の図の場合である．

したがって，$-3\leqq 4a-3<-2$

ゆえに，$0\leqq 4a<1$

よって，$0\leqq a<\frac{1}{4}$

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% 節末問題I1.3.2：（One More）★★
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整数$x$は4の倍数であり，$x$を15で割ったところ，割り切れなかった．そこで$\frac{x}{15}$を計算し，その小数第1位を四捨五入したところ，$4$になった．このとき，整数$x$をすべて求めよ．

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% 節末問題I1.3.2の解答（One More）★★
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$\frac{x}{15}$の小数第1位を四捨五入すると$4$となることから，

\[
3.5\leqq\frac{x}{15}<4.5
\]

各辺を15倍すると，$52.5\leqq x<67.5$

よって，この範囲にある$15$で割り切れない$4$の倍数を求めると，

\[
x=56,64
\]

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% 節末問題I1.3.3：（One More）★★
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(1) 駅から自宅までの道のりは30 kmである．この道のりを，初めは時速5 kmで歩き，途中からは時速10 kmで走ると，掛かった時間は5時間以内であった．時速5 kmで歩いた道のりはどれほどであるか．

(2) $7\%$の食塩水と$10\%$の食塩水がある．$7\%$の食塩水500 gと$10\%$の食塩水を何gか混ぜ合わせて，$8\%$以上$8.5\%$以下の食塩水を作りたい．$10\%$の食塩水を何g以上何g以下混ぜればよいか．

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% 節末問題I1.3.3の解答（One More）★★
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(1) 時速5 kmで歩いた道のりを$x\mathrm{~km}$とすると，歩いた時間は，$\frac{x}{5}$（時間）

また，時速10 kmで走った道のりを$(30-x)\mathrm{~km}$とすると，走った時間は，$\frac{30-x}{10}$（時間）

これらを合わせて5時間以内であるから，

\[
\frac{x}{5}+\frac{30-x}{10}\leqq 5
\]

両辺に10を掛けると，

\begin{align*}
2x+30-x&\leqq 50\\
x&\leqq 20
\end{align*}

よって，時速5 kmで歩いた道のりは，20 km以下である．

(2) $10\%$の食塩水を$x\mathrm{~g}$混ぜるとする．

$7\%$の食塩水500 gに含まれる食塩の量は，$500\times 0.07=35(\mathrm{g})$

$10\%$の食塩水$x\mathrm{~g}$に含まれる食塩の量は，$0.10x(\mathrm{g})$

$7\%$の食塩水500 gに$10\%$の食塩水を$x\mathrm{~g}$混ぜると，食塩水の量は$(500+x)\mathrm{~g}$となるから，その濃度が$8\%$以上$8.5\%$以下になるための条件は，

\[
8\leqq\frac{35+0.10x}{500+x}\times 100\leqq 8.5
\]

各辺に$500+x$を掛けて，

\[
8(500+x)\leqq 3500+10x\leqq 8.5(500+x)
\]

ゆえに，$4000+8x\leqq 3500+10x\leqq 4250+8.5x$

$4000+8x\leqq 3500+10x$より，$2x\geqq 500$

すなわち，$x\geqq 250\cdots(\mathrm{i})$

$3500+10x\leqq 4250+8.5x$より，$1.5x\leqq 750$

すなわち，$x\leqq 500\cdots(\mathrm{ii})$

(i)，(ii)より，$250\leqq x\leqq 500$

よって，$10\%$の食塩水を250 g以上500 g以下だけ混ぜればよい．

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% 節末問題I1.3.4：（One More）★★★
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次の不等式を解け．ただし，$a,b$は定数とする．

(1) $ax>b$

(2) $(a+b)x\leqq a^2-b^2$

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% 節末問題I1.3.4の解答（One More）★★★
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(1)

(i) $a>0$のとき

両辺を$a$で割ると，$x>\frac{b}{a}$

(ii) $a=0$のとき

$0\cdot x>b$となるから，

（ア）$b<0$のとき，解はすべての実数

（イ）$b\geqq 0$のとき，解なし

(iii) $a<0$のとき

両辺を$a$で割ると$x<\frac{b}{a}$

よって，(i)〜(iii)より，求める解は，

\[
\begin{cases}
a>0\text{のとき，}&x>\frac{b}{a}\\
a=0\text{のとき，}&b<0\text{ならば解はすべての実数}\\
&b\geqq 0\text{ならば解なし}\\
a<0\text{のとき，}&x<\frac{b}{a}
\end{cases}
\]

(2) $(a+b)x\leqq a^2-b^2$より，$(a+b)x\leqq(a+b)(a-b)$

(i) $a+b>0$のとき

両辺を$a+b$で割ると$x\leqq a-b$

(ii) $a+b=0$のとき

不等式は，$0\cdot x\leqq 0$となり，$x$の値に関わらず成り立つ．

したがって，解はすべての実数

(iii) $a+b<0$のとき

両辺を$a+b$で割ると$x\geqq a-b$

よって，(i)〜(iii)より，求める解は，

\[
\begin{cases}
a+b>0\text{のとき，}&x\leqq a-b\\
a+b=0\text{のとき，}&\text{解はすべての実数}\\
a+b<0\text{のとき，}&x\geqq a-b
\end{cases}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.5：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の方程式を解け．

(1) $\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}=6$

(2) $x^2+|x+3|+|x-2|=6$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.5の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 方程式の左辺を変形すると，$\sqrt{(x-1)^2}+\sqrt{(x-3)^2}=6$

すなわち，$|x-1|+|x-3|=6$

(i) $x\geqq 3$のとき

$(x-1)+(x-3)=6$より，$x=5$

これは$x\geqq 3$を満たす．

(ii) $1\leqq x<3$のとき

\[
(x-1)-(x-3)=6
\]

これは，$2=6$となり，不適である．

(iii) $x<1$のとき

$-(x-1)-(x-3)=6$より，$x=-1$

これは$x<1$を満たす．

よって，(i)〜(iii)より，$x=5,-1$

(2)

(i) $2\leqq x$のとき

$x^2+(x+3)+(x-2)=6$より，$x^2+2x-5=0$

これを解くと，$x=-1\pm\sqrt{6}$

これらのうち，$2\leqq x$を満たすものはない．

(ii) $-3\leqq x<2$のとき

$x^2+(x+3)-(x-2)=6$より，$x^2=1$

これを解くと，$x=1,-1$

これらは，$-3\leqq x<2$を満たす．

(iii) $x<-3$のとき

$x^2-(x+3)-(x-2)=6$より，$x^2-2x-7=0$

これを解くと，$x=1\pm 2\sqrt{2}$

これらのうち，$x<-3$を満たすものはない．

よって，(i)〜(iii)から，求める解は$x=1,-1$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.6：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の方程式，不等式を解け．

(1) $|2x-3|<3x$

(2) $|x-3|+|x-5|\leqq 5$

(3) $||x-2|+4|=3x$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.6の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1)

(i) $2x-3\geqq 0$，すなわち，$x\geqq\frac{3}{2}$のとき

$2x-3<3x$より，$x>-3$

したがって，$x\geqq\frac{3}{2}$より，$\frac{3}{2}\leqq x$

(ii) $2x-3<0$，すなわち，$x<\frac{3}{2}$のとき

$-(2x-3)<3x$より，$5x>3$

したがって，$x>\frac{3}{5}$

ゆえに，$x<\frac{3}{2}$より，$\frac{3}{5}<x<\frac{3}{2}$

よって，(i)，(ii)より，$\frac{3}{5}<x$

(2)

(i) $x\geqq 5$のとき

$(x-3)+(x-5)\leqq 5$より，$x\leqq\frac{13}{2}$

したがって，$x\geqq 5$より，$5\leqq x\leqq\frac{13}{2}$

(ii) $3\leqq x<5$のとき

$(x-3)-(x-5)\leqq 5$より，$2\leqq 5$となり，これは成り立っている．

したがって，$3\leqq x<5$

(iii) $x<3$のとき

$-(x-3)-(x-5)\leqq 5$より，$x\geqq\frac{3}{2}$

したがって，$x<3$より，$\frac{3}{2}\leqq x<3$

よって，(i)〜(iii)より，$\frac{3}{2}\leqq x\leqq\frac{13}{2}$

(3)

\begin{align*}
||x-2|+4|&=\begin{cases}|x-2+4|&(x\geqq 2)\\
|-(x-2)+4|&(x<2)\end{cases}\\
&=\begin{cases}|x+2|&(x\geqq 2)\\
|-x+6|&(x<2)\end{cases}\\
&=\begin{cases}x+2&(x\geqq 2)\\
-x+6&(x<2)\end{cases}
\end{align*}

(i) $x\geqq 2$のとき

$x+2=3x$より，$x=\frac{2}{2}=1$

これは，$x\geqq 2$を満たさない．

(ii) $x<2$のとき

$-x+6=3x$より，$6=4x$より，$x=\frac{6}{4}=\frac{3}{2}$

これは，$x<2$を満たす．

よって，(i)，(ii)より，$x=\frac{3}{2}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.1：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の式を展開せよ．

\[
(x+y+z)^2-(y+z-x)^2+(z+x-y)^2-(x+y-z)^2
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.1の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{align*}
&(x+y+z)^2-(y+z-x)^2+(z+x-y)^2-(x+y-z)^2\\
=&\{(x+y+z)+(y+z-x)\}\{(x+y+z)-(y+z-x)\}\\
&+\{(z+x-y)+(x+y-z)\}\{(z+x-y)-(x+y-z)\}\\
=&2(y+z)\cdot 2x+2x\cdot 2(z-y)\\
=&4xy+4xz+4xz-4xy=8xz
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.2：（One More）★★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の式を因数分解せよ．

(1) $a^3(b-c)+b^3(c-a)+c^3(a-b)$

(2) $a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.2の解答（One More）★★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1)

\begin{align*}
&a^3(b-c)+b^3(c-a)+c^3(a-b)\\
=&(b-c)a^3-(b^3-c^3)a+bc(b^2-c^2)\\
=&(b-c)\{a^3-(b^2+bc+c^2)a+bc(b+c)\}\\
=&(b-c)\{(c-a)b^2+c(c-a)b-a(c^2-a^2)\}\\
=&(b-c)(c-a)\{b^2+c\cdot b-a(c+a)\}\\
=&(b-c)(c-a)\{(b-a)c+(b^2-a^2)\}\\
=&(b-c)(c-a)(b-a)\{c+(b+a)\}\\
=&-(a-b)(b-c)(c-a)(a+b+c)
\end{align*}

(2)

\begin{align*}
&a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2\\
=&a^4-2(b^2+c^2)a^2+b^4-2b^2c^2+c^4\\
=&a^4-2(b^2+c^2)a^2+(b^2-c^2)^2\\
=&a^4-2(b^2+c^2)a^2+\{(b+c)(b-c)\}^2\\
=&a^4-\{(b+c)^2+(b-c)^2\}a^2+(b+c)^2(b-c)^2\\
=&\{a^2-(b+c)^2\}\{a^2-(b-c)^2\}\\
=&\{a+(b+c)\}\{a-(b+c)\}\{a+(b-c)\}\{a-(b-c)\}\\
=&(a+b+c)(a-b-c)(a+b-c)(a-b+c)
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.3：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$x+y+z=0$のとき，$x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y})$の値を求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.3の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$x+y+z=0$より，$z=-(x+y)$

これより，与えられた式を$z$について整理すると，

\begin{align*}
&x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y})\\
&=(\frac{1}{x}+\frac{1}{y})z+\frac{x+y}{z}+\frac{x}{y}+\frac{y}{x}\\
&=-(\frac{1}{x}+\frac{1}{y})(x+y)+\frac{x+y}{-(x+y)}+\frac{x}{y}+\frac{y}{x}\\
&=-(1+\frac{y}{x}+\frac{x}{y}+1)-1+\frac{x}{y}+\frac{y}{x}=-3
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.3の別解
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$x+y+z=0$より，$y+z=-x,z+x=-y,x+y=-z$

よって，

\begin{align*}
&x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y})\\
&=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\\
&=\frac{-x}{x}+\frac{-y}{y}+\frac{-z}{z}=-3
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.4：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

不等式$|ax+2|\leqq b$の解が$-2\leqq x\leqq 4$のとき$a,b$の値を求めよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.4の解答（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$|ax+2|\leqq b$より，$-b\leqq ax+2\leqq b$

よって，$-b-2\leqq ax\leqq b-2$

(i) $a>0$のとき

$\frac{-b-2}{a}\leqq x\leqq\frac{b-2}{a}$より，$\frac{-b-2}{a}=-2,\frac{b-2}{a}=4$

これを解いて，$a=-2,b=-6$

これは，$a>0$を満たさないので不適である．

(ii) $a=0$のとき

このとき，解は$-2\leqq x\leqq 4$とはならないので不適である．

(iii) $a<0$のとき

$\frac{b-2}{a}\leqq x\leqq\frac{-b-2}{a}$より，$\frac{b-2}{a}=-2,\frac{-b-2}{a}=4$

これを解いて，$a=-2,b=6$

これは，$a<0$を満たす．

(i)〜(iii)より，$a=-2,b=6$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.5：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$x=2a-1$のとき，$\sqrt{x^2+8a}+\sqrt{a^2-x}$を簡単にせよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.5の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$x=2a-1$を与えられた式に代入すると，

\begin{align*}
&\sqrt{(2a-1)^2+8a}+\sqrt{a^2-(2a-1)}\\
&=\sqrt{4a^2+4a+1}+\sqrt{a^2-2a+1}\\
&=\sqrt{(2a+1)^2}+\sqrt{(a-1)^2}\\
&=|2a+1|+|a-1|
\end{align*}

よって，

\begin{align*}
|2a+1|+|a-1|&=\begin{cases}(2a+1)+(a-1)&(1\leqq a)\\
(2a+1)-(a-1)&(-\frac{1}{2}\leqq a<1)\\
-(2a+1)-(a-1)&(a<-\frac{1}{2})\end{cases}\\
&=\begin{cases}3a&(1\leqq a)\\
a+2&(-\frac{1}{2}\leqq a<1)\\
-3a&(a<-\frac{1}{2})\end{cases}
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.1.1の側注：（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
先に$P$を求める．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.1.2の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
直接すべての項を展開してもよいが，計算に手間が掛かる．そこで，掛けて$x^5,x^3$になるものだけを取り出して考える．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.1.3の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$(A+B)(A-B)=A^2-B^2$

% 側注2
$(A+B)(A-B)=A^2-B^2$

% 側注3
輪環の順に整理するとよい．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.1.4の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
先に，(i)のように$2$をくくり出して，$2y^2+y-1$をたすき掛けを用いて因数分解する．次に，(ii)のように波括弧内の全体を$x$の2次式と考えて，たすき掛けを用いて因数分解する．

% 側注2
$A^2-B^2=(A+B)(A-B)$

% 側注3
$x+y=A$とおくと，

\begin{align*}
&A^3+z^3\\
=&(A+z)(A^2-Az+z^2)
\end{align*}

% 側注4
$x^3=A$とおくと，

\[
A^2-1=(A+1)(A-1)
\]

なお，$x^2=A$とおいて，3次の乗法公式を用いてもよい．

% 側注5
$a^3=A$とおくと，

\begin{align*}
&A^2-7A-8\\
=&(A+1)(A-8)
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.1.5の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
\begin{align*}
&-(x+y-2z)^3\\
=&(-x-y+2z)^3
\end{align*}

% 側注2
$a+b+c=0$

% 側注3
$4x^4$と$16y^4$より，$(2x^2+4y^2)^2$または$(2x^2-4y^2)^2$を作ることを考える．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.1の側注：（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$0.\dot{1}\dot{5}$を既約分数で表す．

% 側注2
$0.\dot{5}\dot{4}$を既約分数で表す．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.2の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\sqrt{a^2}=|a|$を利用する．

\[
|a|=\{\begin{array}{rr}a&(a\geqq 0)\\-a&(a<0)\end{array}
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.3の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
例えば$\frac{1}{\sqrt{1}+\sqrt{2}}$は，

\begin{align*}
&\frac{1}{\sqrt{1}+\sqrt{2}}\\
=&\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}\\
=&\sqrt{2}-\sqrt{1}
\end{align*}

となる．他の項も同様に分母の有理化を行う．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.4の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$(\sqrt{2})^2+(\sqrt{3})^2=(\sqrt{5})^2$であることに着目して，第1項と第2項の分母を有理化する．

% 側注2
第1項と第2項を通分する．

% 側注3
分母，分子に$\sqrt{15}-3$を掛けて，分母を有理化する．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.5の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
先に，$\sqrt{4+2\sqrt{3}}$を簡単な形にする．和が$4$，積が$3$になる2つの数は，$3$と$1$である．

% 側注2
$\sqrt{8+4\sqrt{3}}$を簡単な形にする．$4\sqrt{3}=2\sqrt{12}$であり，和が$8$，積が$12$になる2つの数は，$6$と$2$である．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.6の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
(i)を移項して，$a+b=3-c,b+c=3-a,c+a=3-b$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.2.7の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
分母を有理化する．

% 側注2
$3^2<11<4^2$より，$3<\sqrt{11}<4$

% 側注3
$\frac{7}{5}=1.4,\frac{8}{5}=1.6$

% 側注4
小数部分は，

$(\text{もとの数})-(\text{整数部分})$

で求められる．

% 側注5
$6^2<(2\sqrt{11})^2<7^2$より，$6<2\sqrt{11}<7$

% 側注6
\begin{align*}
&(\text{もとの数})\\
=&(\text{整数部分})+(\text{小数部分})
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.1の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
等号を含むか否かに注意すること．例えば$4a-3=-3$のとき，共通範囲の不等式は$-3<x<\frac{6}{5}$となり，解に含まれる整数は4つとなる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.2の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
等号を含むか否かに注意すること．

% 側注2
$60$は$15$で割り切れる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.3の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
求めるものを$x$とおく．

% 側注2
$(\text{時間})=\frac{(\text{道のり})}{(\text{速さ})}$

% 側注3
求めるものを$x$とおく．

% 側注4
\begin{align*}
&(\text{食塩の質量})\\
=&(\text{食塩水の質量})\times(\text{濃度})
\end{align*}

% 側注5
$(\text{濃度})=\frac{(\text{食塩の質量})}{(\text{食塩水の質量})}$

% 側注6
$500+x$は正であるので，不等号の向きは変わらず，1次不等式の形に変形することができる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.4の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$b$の符号によって，さらに$b$の場合分けが必要となる．

% 側注2
$a<0$より，不等号の向きが変わる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.5の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$||$内の式について，$x-1=0$の解は$x=1$であり，$x-3=0$の解は$x=3$となる．これより，$x<1,1\leqq x<3,3\leqq x$の場合に分ける．

% 側注2
$2^2<(\sqrt{6})^2<3^2$より，$2<\sqrt{6}<3$

% 側注3
$2^2<(2\sqrt{2})^2<3^2$より，$2<2\sqrt{2}<3$

% 側注4
$\sqrt{a^2}=|a|$を利用する．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 節末問題I1.3.6の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$x\geqq 5$のとき，$|x-3|=x-3,|x-5|=x-5$

% 側注2
$3\leqq x<5$のとき，$|x-3|=x-3,|x-5|=-(x-5)$

% 側注3
$x<3$のとき，$|x-3|=-(x-3),|x-5|=-(x-5)$

% 側注4
(i)〜(iii)の範囲を数直線で表すと次のようになる．

% 側注5
先に，$x-2$の絶対値記号を外し，$x-2\geqq 0$と$x-2<0$で場合分けをする．

% 側注6
$x\geqq 2$のとき，$x+2>0$

$x<2$のとき，$-x+6>0$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.1の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
\begin{align*}
&A^2-B^2+C^2-D^2\\
&=(A+B)(A-B)\\
&+(C+D)(C-D)
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.2の側注：（One More）★★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$a$について整理する．

% 側注2
$b-c$をくくり出す．

% 側注3
波括弧内を$b$について整理する．

% 側注4
$c-a$をくくり出す．

% 側注5
波括弧内を$c$について整理する．

% 側注6
$b-a$をくくり出す．

% 側注7
$a$について整理する．

% 側注8
$a^2-2ab+b^2=(a-b)^2$

% 側注9
次の式は公式として覚えてもよい．

\begin{align*}
&(x+y)^2+(x-y)^2\\
=&2(x^2+y^2)
\end{align*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.3の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
展開して，分母が同じものをまとめる．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.4の側注：（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\frac{-b-2}{a}=-2,\frac{b-2}{a}=4$の各辺を足し合わせるなどして，連立させて$a$の値を求める．

% 側注2
$\frac{b-2}{a}=-2,\frac{-b-2}{a}=4$の各辺を足し合わせるなどして，連立させて$a$の値を求める．

% 側注3
$|X|\leqq c(c\geqq 0)$の解は，$-c\leqq X\leqq c$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 章末問題I1.5の側注：（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$||$内の式について，$2a+1=0$の解は$a=-\frac{1}{2}$であり，$a-1=0$の解は$a=1$となる．これより，$a<-\frac{1}{2},-\frac{1}{2}\leqq a<1,1\leqq a$の場合に分ける．