đkxđ: x≠\(\pm\)2
pt <=> \(\left(x+1\right)\left(x^2-4\right)=x-2\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x+1\right)\left(x+2\right)-1\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+1\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x^2+3x+1=0\end{matrix}\right.\)
+) x-2=0 => x = 2(ktm)
+) \(x^2+3x+1=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\sqrt{\dfrac{5}{4}}\\x+\dfrac{3}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-3}{2}\\x=\dfrac{-\sqrt{5}-3}{2}\end{matrix}\right.\)(t/m)
vậy...........
