a: \(\Leftrightarrow\dfrac{x^2+2x+1-2x+4}{\left(x-2\right)\left(x+1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2+10=x^2-x-2\)
\(\Leftrightarrow x^2+x+12=0\)
hay \(x\in\varnothing\)
b: =>2-x=3 hoặc 2-x=-3
=>x=-1 hoặc x=5
a, đk x khác -1 ; 2
\(\dfrac{2\left(x+1\right)^2-4\left(x-2\right)}{2\left(x+1\right)\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+1\right)}{2\left(x+1\right)\left(x-2\right)}\Rightarrow2x^2+4x+2-4x+8=x^2-x-2\)
\(\Leftrightarrow x^2+x+12=0\)( vô lí ) Vì x^2 + x + 12 > 0
b, \(\left|2-x\right|=3\Leftrightarrow\left[{}\begin{matrix}2-x=3\\2-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\end{matrix}\right.\)
c, \(\left|2-x\right|=x-2\)đk x > = 2
\(\Leftrightarrow2-x=x-2\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)
