Đáp án:
a,b) $S=\varnothing $
c) $x=\frac{11}{12}$
d)$x=\frac{5}{11}$
e)$x=4,x=\frac{1}{3}$
f)$x=0$
Giải thích các bước giải:
$a) \frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\\
DK: x\neq -1\\
\Rightarrow \frac{1-x}{x+1}+\frac{3(x+1)}{x+1}=\frac{2x+3}{x+1}\\
\Leftrightarrow 1-x+3x+3=2x+3\\
\Leftrightarrow 2x+4-2x=3\\
\Leftrightarrow 0x=-1\\
S=\varnothing \\
b) \frac{(x+2)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\\
DK: x \neq \frac{3}{2}\\
\Rightarrow \frac{(x+2)^2}{2x-3}-\frac{2x-3}{2x=3}=\frac{x^2+10}{2x-3}\\
\Leftrightarrow x^2+4x+4-2x+3=x^2+10\\
\Leftrightarrow 2x=3\\
\Leftrightarrow x=\frac{3}{2}(loai)\\
S=\varnothing \\
c) \frac{5x-2}{2-2x}+\frac{2x-1}{2}=1-\frac{x^2+x-3}{1-x}\\
DK: x \neq 1\\
\Rightarrow \frac{5x-2}{2(1-x)}+\frac{(2x-1)(1-x)}{2(1-x)}=\frac{2-2x}{2(1-x)}-\frac{2(x^2+x-3)}{2(1-x)}\\
\Leftrightarrow 5x-2+2x-2x^2-1+x=2-2x-2x^2-2x+6\\
\Leftrightarrow 8x-3=-4x+8\\
\Leftrightarrow 12x=11\\
\Leftrightarrow x=\frac{11}{12}\\$
$d)\frac{5-2x}{3}+\frac{(x-1)(x+1)}{3x-1}=\frac{(x+2)(1-3x)}{9x-3}\\
DK: x \neq \frac{1}{3}\\
\Rightarrow \frac{(5-2x)(3x-1)}{3(3x-1)}+\frac{3(x-1)(x+1)}{3(3x-1)}=\frac{(x+2)(1-3x)}{3(3x-1)}\\
\Leftrightarrow 15x-5-6x^2+2x+3(x^2-1)=x+2-3x^2-6x\\
\Leftrightarrow 17x-5-6x^2+3x^2-3=-3x^2-5x+2\\
\Leftrightarrow 17x-8-3x^2+3x^2+5x=2\\
\Leftrightarrow 22x=10\\
\Leftrightarrow x=\frac{5}{11}\\$
$e) \frac{2x+1}{x-1}=\frac{5(x-1)}{x+1}\\
DK: x \neq 1, x\neq -1\\
\Rightarrow \frac{(2x+1)(x+1)}{(x+1)(x-1)}=\frac{5(x-1)^2}{(x-1)(x+1)}\\
\Leftrightarrow 2x^2+2x+x+1=5(x^2-2x+1)$
$\Leftrightarrow 2x^2+3x+1=5x^2-10x+5$
$\Leftrightarrow 3x^2-13x+4=0$
$\Leftrightarrow 3x^2-12x-x+4=0$
$\Leftrightarrow 3x(x-4)-(x-4)=0$
$\Leftrightarrow (x-4)(3x-1)=0$
$\Leftrightarrow {\left [\begin{aligned}x-4=0\\3x-1=0\end{aligned}\right.}$
$\Leftrightarrow {\left[\begin{aligned}x=4\\x=\frac{1}{3}\end{aligned}\right.}$
$f) \frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\\
DK: x \neq 1\\
\Rightarrow \frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4(x-1)}{x^3-1}\\
\Leftrightarrow x^2+x+1+2x^2-5=4x-4\\
\Leftrightarrow 3x^2-3x=0\\
\Leftrightarrow 3x(x-1)=0\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x-1=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=0\\x=1(loai)\end{aligned}\right.}$
