Giải thích các bước giải:
$o)\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2$ $(x\neq\{-1;0\})$
$⇒\dfrac{x(x+3)}{x(x+1)}+\dfrac{(x-2)(x+1)}{x(x+1)}=\dfrac{2x(x+1)}{x(x+1)}$
$⇒x^2+3x+x^2-x-2=2x^2+2x$
$⇒2x-2=2x$
$⇒2=0_\text{(vô lí)}$
Vậy phương trình vô nghiệm
$p)\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x(x-2)}$ $(x\neq\{0;2\})$
$⇒\dfrac{x(x+2)}{x(x-2)}-\dfrac{x-2}{x(x-2)}=\dfrac{2}{x(x-2)}$
$⇒x^2+2x-x+2=2$
$⇒x^2+x=0$
$⇒x(x+1)=0$
$⇒\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.⇒\left[ \begin{array}{l}x=0_{(ktm)}\\x=-1_{(tm)}\end{array} \right.$
Vậy $x=-1$
$q)\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{(x+1)(x-2)}$ $(x\neq\{-1;2\})$
$⇒\dfrac{2(x-2)}{(x+1)(x-2)}-\dfrac{x+1}{(x+1)(x-2)}=\dfrac{3x-11}{(x+1)(x-2)}$
$⇒2x-4-x-1=3x-11$
$⇒x-5=3x-11$
$⇒-2x=-6$
$⇒x=3_{(tm)}$
Vậy $x=3$
$r)\dfrac{x-2}{x+2}+\dfrac{3}{x-2}=\dfrac{x^2-11}{x^2-4}$ $(x\neq±2)$
$⇒\dfrac{(x-2)^2}{(x+2)(x-2)}+\dfrac{3(x+2)}{(x+2)(x-2)}=\dfrac{x^2-11}{x^2-4}$
$⇒x^2-4x+4+3x+6=x^2-11$
$⇒-x+10=-11$
$⇒x=21_{(tm)}$
Vậy $x=21$
$s)\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{(2x-1)(2x+1}$ $(x\neq±\dfrac{1}{2})$
$\text{$⇒\dfrac{2x(2x+1)}{(2x+1)(2x-1)}+\dfrac{x(2x-1)}{(2x+1)(2x-1)}=\dfrac{(2x+1)(2x-1)}{(2x+1)(2x-1)}+\dfrac{4}{(2x+1)(2x-1)}$}$
$⇒4x^2+2x+2x^2-x=4x^2-1+4$
$⇒2x^2+x=3$
$⇒2x^2+x-3=0$
$⇒2x^2-2x+3x-3=0$
$⇒2x(x-1)+3(x-1)=0$
$⇒(2x+3)(x-1)=0$
$⇒\left[ \begin{array}{l}2x+3=0\\x-1=0\end{array} \right.⇒\left[ \begin{array}{l}x=-\dfrac{3}{2}_{(tm)}\\x=1_{(tm)}\end{array} \right.$
Vậy $x∈\{-\dfrac{3}{2};1\}$
