Câu hỏi của DRACULA

Question

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

Rồng Đom Đóm · Accepted Answer

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

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

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

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

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

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

$\Leftrightarrow\left[{}\begin{matrix}x=-1\x=-2\left(loai\right)\end{matrix}\right.$Câu trả lời: $\left(x+1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2$

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

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

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

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

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

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

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