ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow x^3+3x\left(x+2\right)-4\left(x+2\right)\sqrt{x+2}=0\)
Đặt \(\sqrt{x+2}=y\ge0\) pt trở thành:
\(x^3+3xy^2-4y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+4y^2\right)=0\)
\(\Leftrightarrow x=y\Leftrightarrow\sqrt{x+2}=x\) (\(x\ge0\))
\(\Leftrightarrow x^2=x+2\Leftrightarrow x=2\)
\(ĐKXĐ:x\ge-2\)
\(\Leftrightarrow x^3+3x^2+6x-4x\sqrt{x+2}-8\sqrt{x+2}=0\Leftrightarrow4x^2-4x\sqrt{x+2}+8x-8\sqrt{x+2}+x^3-x\left(x+2\right)=0\Leftrightarrow4x\left(x-\sqrt{x+2}\right)+8\left(x-\sqrt{x+2}\right)+x\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)=0\)\(\Leftrightarrow\left(x-\sqrt{x+2}\right)\left(x^2+x\sqrt{x+2}+4x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+2}=0\left(1\right)\\x^2+x\sqrt{x+2}+4x+8=0\left(2\right)\end{matrix}\right.\) Từ (1) \(\Rightarrow x=\sqrt{x+2}\left(x\ge0\right)\Rightarrow x^2=x+2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(L\right)\end{matrix}\right.\) Từ (2) \(\Rightarrow x^2+x\sqrt{x+2}+4x+8\ge\left(-2\right)^2+\left(-2\right)\sqrt{-2+2}+4\left(-2\right)+8=4>0\) \(\Rightarrow\) ko có x
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