数学学習の記録 126.1 数・式の計算・方程式不等式 (数学読本) 第2章問18,19

"数・式の計算・方程式不等式 (数学読本)"の第2章問18,19を解いてみる。

問18

(1)

(2)

(3)

$x^{2}-2x-2$

問19

(1)

$\frac{ac-bc+ab-ac+bc-ab}{abc}=0$

(2)

$\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}$

(3)

$\frac{x+2-4}{x-2}=1$

(4)

$\frac{x-1-2x}{(x+1)(x-1)}=\frac{1}{1-x}$

(5)

$\frac{x^{2}+3x+2-2x}{x+1}-\frac{3x^{2}+4}{x(x+1)}\\ =\frac{x^{2}+x+2}{x+1}-\frac{3x^{2}+4}{x(x+1)}\\ =\frac{x^{3}+x^{2}+2x-3x^{2}-4}{x(x+1)}$

$=\frac{x^{3}-2x^{2}+2x-4}{x(x+1)}\\ =\frac{(x^{2}+2)(x-2)}{x(x+1)}$

(6)

$\frac{x^{2}-x-2-x^{2}+x-1+x^{2}+x+3}{x^{3}+1}\\ =\frac{x(x+1)}{(x+1)(x^{2}-x+1)}\\ =\frac{x}{x^{2}-x+1}$

(7)

$\frac{1}{(x-1)(x-2)}+\frac{2}{(x-1)(2x+1)}-\frac{3}{(x-2)(2x+1)}\\ =\frac{2x+1+2x-4-3x+3}{(x-1)(x-2)(2x+1)}$

$=\frac{x}{(x-1)(x-2)(2x+1)}$

(8)

$\frac{(x+1)(x+2)(x+3)-x(x+2)(x+3)-x(x+1)(x+3)+x(x+1)(x+2)}{x(x+1)(x+2)(x+3)}\\ =\frac{(x+2)(x+3)-x(x+1)}{x(x+1)(x+2)(x+3)}$

$=\frac{2(2x+3)}{x(x+1)(x+2)(x+3)}$

(9)

$=\frac{a+1+a-1}{a^{2}-1}+\frac{2a}{a^{2}+1}+\frac{4a^{3}}{a^{4}+1}\\ =\frac{2a^{3}+2a+2a^{3}-2a}{a^{4}-1}+\frac{4a^{3}}{a^{4}+1}$

$=\frac{4a^{3}(a^{4}+1)+4a^{7}-4a^{3}}{a^{8}-1}\\ =\frac{8a^{7}}{a^{8}-1}$

(10)

$\frac{(x+2)(x+3)+x(x+3)+x(x+1)}{x(x+1)(x+2)(x+3)}\\ =\frac{3(x^{3}+3x+2)}{x(x+1)(x+2)(x+3)}$

$=\frac{3}{x(x+3)}$

(11)

$\frac{-a(b-c)-b(c-a)-c(a-b)}{(a-b)(b-c)(c-a)}\\ =0$

(12)

$\frac{-(b+1)(c+1)(b-c)-(a+1)(c+1)(c-a)-(a+1)(b+1)(a-b)}{(a-b)(b-c)(c-a)(a+1)(b+1)(c+1)}$

$=\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)(a+1)(b+1)(c+1)}\\ =\frac{1}{(a+1)(b+1)(c+1)}$

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