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mọi người giúp e giải pt trên với ạ

Nguyễn Việt Lâm · Accepted Answer

$\Leftrightarrow4x^2=x+2+2\sqrt{x+2}+1$

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

$\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}+1=2x\\sqrt{x+2}+1=-2x\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}=2x-1\left(x\ge\dfrac{1}{2}\right)\\sqrt{x+2}=-2x-1\left(x\le-\dfrac{1}{2}\right)\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x+2=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\x+2=4x^2+4x+1\left(x\le-\dfrac{1}{2}\right)\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{41}}{8}\x=-1\end{matrix}\right.$

Câu trả lời:

$\Leftrightarrow4x^2=x+2+2\sqrt{x+2}+1$

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

$\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}+1=2x\\sqrt{x+2}+1=-2x\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}=2x-1\left(x\ge\dfrac{1}{2}\right)\\sqrt{x+2}=-2x-1\left(x\le-\dfrac{1}{2}\right)\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x+2=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\x+2=4x^2+4x+1\left(x\le-\dfrac{1}{2}\right)\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{41}}{8}\x=-1\end{matrix}\right.$