Đáp án:
Giải thích các bước giải:
$38)$
$a)ĐKXĐ:x\neq -1$
$\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}$
$\Rightarrow 1-x+3(x+1)=2x+3$
$\Leftrightarrow 1-x+3x+3=2x+3$
$\Leftrightarrow 2x+4=2x+3$
$\Leftrightarrow 0x=1$(vô lí)
Vậy phương trình vô nghiệm
$b)ĐKXĐ:x\neq \dfrac{3}{2}$
$\dfrac{(x+2)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}$
$\Rightarrow (x+2)^2-(2x-3)=x^2+10$
$\Leftrightarrow x^2+4x-2x+3=x^2+10$
$\Leftrightarrow 2x=7$
$\Leftrightarrow x=\dfrac{7}{2}(tm)$
Vậy ...
$c)ĐKXĐ:x\neq 1$
$\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2+x-3}{1-x}$
$\Leftrightarrow \dfrac{5x-2}{2(1-x}+\dfrac{2x-1}{2}=1-\dfrac{x^2+x-3}{1-x}$
$\Rightarrow 5x-2+(2x-1)(1-x)=2(1-x)-2(x^2+x-2)$
$\Leftrightarrow 5x-2+3x-2x^2-1=2-2x-2x^2-2x+4$
$\Leftrightarrow -2x^2+8x-3=-2x^2-4x+6$
$\Leftrightarrow 12x=9$
$\Leftrightarrow x=\dfrac{3}{4}(tm)$
Vậy ...
$d)ĐKXĐ:x\neq 1;x\neq \dfrac{1}{3}$
$\Leftrightarrow \dfrac{5-2x}{2(1-x)}+\dfrac{(x-1)(x+1)}{3x-1}=\dfrac{(x-2)(1-3x)}{3(3x-1)}$
$\Rightarrow 3(3x-1)(5-2x)+6(x-1)(x+1)(1-x)=2(x-2)(1-x)(1-3x)$
...
$41)$
$a)ĐKXĐ:x\neq 1;x\neq -1$
$\Rightarrow (2x+1)(x+1)=5(x-1)^2$
$\Leftrightarrow 2x^2+3x+1=5x^2-10x+5$
$\Leftrightarrow 3x^2-13x+4=0$
$\Leftrightarrow x=\dfrac{1}{3}(ktm)$ hoặc $x=4(tm)$
Vậy...
