Đáp án:
$a) $$\dfrac{1}{x-1}$ $+ 1 =$ $\dfrac{x}{x-2}$ (điều kiện $x$$\neq$ $1; x$$\neq$ $2$)
$⇔$ $\dfrac{(x-2)+(x-1)(x-2)}{(x-1)(x-2)}$ $=$ $\dfrac{x(x-1)}{(x-1)(x-2)}$
$⇔ (x-2)+(x-1)(x-2) = x(x-1)$
$⇔ x-2+x²-2x-x+2-x²+x=0$
$⇔ -x+0=0$
$⇔ x=0$ (nhận)
Vậy $S=${$0$}
$b)$ $\dfrac{5}{x+7}$ $+$ $\dfrac{8}{2x+14}$ $=$ $\dfrac{3}{2}$ (điều kiện $x$$\neq$ $-7$)
$⇔$ $\dfrac{5}{x+7}$ $+$ $\dfrac{8}{2(x+7)}$ $=$ $\dfrac{3}{2}$
$⇔$ $\dfrac{5}{x+7}$ $+$ $\dfrac{4}{x+7}$ $=$ $\dfrac{3}{2}$
$⇔ 5.2+4.2=3(x+7)$
$⇔ 10+8=3x+21$
$⇔ 18-3x-21=0$
$⇔ -3-3x=0$
$⇔ 3x=-3$
$⇔ x=-1$ (nhận)
Vậy $S=${$-1$}
BẠN THAM KHẢO NHA!!!
