Đáp án:a)$x=±\sqrt{7}$
b)$x=±\frac{1}{2}$
c)$x=\frac{3}{26}$
d)$x=\frac{3}{5}$
e)x=3
g)\(\left[ \begin{array}{l}x=1\\x=\frac{-1}{4}\end{array} \right.\)
h)$x= -\frac{5}{6}$
Giải thích các bước giải:
a, ĐK ; x$\neq$ 1,x$\neq$ -2
pt⇔$\frac{2}{x-1}$= $\frac{x+3}{x+2}$
⇔$2x+4=x^{2}+3x-x-3 ⇔ x^{2}=7 ⇔ x=±\sqrt{7}$
b, $x\neq ±1$
Pt⇔$\frac{(x+1)²-(x-1)²-4}{(x-1)(x+1)}$ =0
⇔$4x^{2}.4-4=0 ⇔ x^{2}=\frac{1}{4} ⇔x=±\frac{1}{2}$
c, đk: x $\neq ±\frac{1}{4}$
Pt⇔ $\frac{2(1-4x)-8-6x-3(4x+1)}{(4x+1)(1-4x}$
⇔$26x=3 ⇔x=\frac{3}{26}$
d,đk: x$\neq \frac{1}{5}$ ,$x\neq \frac{3}{5}$
Pt⇔$\frac{3x(3-5x)(+2(5x-1)-4}{(3-5x)(5x-1)}$
⇔$5x=3 ⇔x=\frac{3}{5}( ktm)$
e, $x\neq ±1$
Pt⇔$\frac{2x+19-85+15(x+1)}{5(x²-1)}$
$⇔17x=51 ⇔x=3$ ( tm)
g, $x\neq 1$
Pt ⇔$\frac{x²+x+1-3x²-2x(x-1)}{x³-1} =0$
$⇔-4x²+3x+1=0$ ⇔\(\left[ \begin{array}{l}x=1(ktm)\\x=\frac{-1}{4}(tm)\end{array} \right.\)
h, đk : $x \neq\frac{-1}{2}, x\neq\frac{-3}{2}$
Pt⇔ $\frac{3(2x+3)-6(2x+1)-8}{4x²+8x+3} =0$
⇔$ -6x=5 ⇔ x= -\frac{5}{6}( tm)$
