Đáp án:a)x=8
b)$x=\frac{11}{8}$
c)$x=3$
d)$x=\frac{-5}{3}$
e)x=-1
Giải thích các bước giải:
$a)\frac{4x-3}{x-5}=\frac{29}{3} (đk: x\neq5)$
$⇔\frac{4x-3}{x-5}-\frac{29}{3}=0$
$⇒3(4x-3)-29(x-5)=0⇔ 12x-9-29x+145=0$
$⇔-17x=-136⇔ x=8 (tm)$
Vậy x=8
$b)\frac{2x-1}{5-3x}=2 (đk: x\neq\frac{5}{3})$
$⇔\frac{2x-1}{5-3x}-2=0$
$⇒2x-1-2(5-3x)=0⇔ 2x-1-10+6x=0$
$⇔8x=11⇔ x=\frac{11}{8} (tm)$
Vậy $x=\frac{11}{8}$
$c)\frac{4x-5}{x-1}=2+\frac{x}{x-1} (đk: x\neq1)$
$⇔\frac{4x-5}{x-1}-2-\frac{x}{x-1}=0$
$⇒4x-5-2(x-1)-x=0⇔ 4x-5-2x+2-x=0$
$⇔x=3 (tm)$
Vậy x=3
$d)\frac{2x+5}{2x}-\frac{x}{x+5}=0 (đk: x\neq0; x\neq-5)$
$⇒ (2x+5)(x+5)-x·2x=0$
$⇔2x^{2}+15x+25-2x^{2}=0$
$⇔15x+25=0⇔ x=\frac{-5}{3} (tm)$
Vậy $x=\frac{-5}{3}$
$e)\frac{12}{1-9x^{2}}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x} (x\neq±\frac{1}{3})$
$⇔\frac{12}{(1-3x)(1+3x)}-\frac{1-3x}{1+3x}+\frac{1+3x}{1-3x}=0$
$⇒12-(1-3x)(1-3x)+(1+3x)(1+3x)=0$
$⇔12-9x^{2}+6x-1+9x^{2}+6x+1=0$
$⇔12x+12=0⇔ x=-1 (tm)$
Vậy x=-1
