Question

giải pt: \(\left(x+1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)

Answer 1

\(\left(x+1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)

\(\Leftrightarrow\left(x+1\right)\sqrt{2\left(x^2+4\right)}=\left(x+1\right)\left(x-2\right)\)

\(\Leftrightarrow\left(x+1\right)\left[\sqrt{2\left(x^2+4\right)}-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x^2+4\right)}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2+8=x^2-4x+4\left(ĐK:x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+4x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(loai\right)\end{matrix}\right.\)

Answer 2