Đáp án câu 1 đề 1 kiểm tra học kì II toán 8

1. a) $3x$ - $11$ = $x + 7$

$\Leftrightarrow $ $3x$ - $x$ = $11+7$

$\Leftrightarrow $ $2x$ = $18$

$\Leftrightarrow $ $x$ = $9$

S = {$9$}

b) $2x$$(x-3)$ = $x-3$

$\Leftrightarrow $ $2x$$(x-3)$ - $x-3$ = $0$

$\Leftrightarrow $ $(x-3)$$(2x-1)$ = $0$

$\Leftrightarrow $ $x$ = $3$ hoặc $x$ = $\frac{1}{2}$

S = {$3$;$\frac{1}{2}$}

c) $\frac{x+2}{x-2}$ - $\frac{5}{x}$ = $\frac{8}{x^{2}-2x}$

Đk: $x$ $\neq $ $0$; $x$ $\neq $ $2$

$\Leftrightarrow $ $\frac{x+2}{x-2}$ - $\frac{5}{x}$ = $\frac{8}{x(x-2)}$

$\Leftrightarrow $ $\frac{x(x+2)}{x(x-2)}$ - $\frac{5(x-2)}{x(x-2)}$ = $\frac{8}{x(x-2)}$ 

$\Leftrightarrow $ $\frac{x^{2} + 2x - 5x +10 -8}{x(x-2)}$ = $0$

$\Leftrightarrow $ $\frac{x^{2} - 3x + 2}{x(x-2)}$ = $0$

$\Leftrightarrow $ $\frac{(x-1)(x-2)}{x(x-2)}$ = $0$

$\Leftrightarrow $ $\frac{(x-1)}{x}$ = $0$

$\Leftrightarrow $ $x$ = $1$

S = {1}