ĐK: \(x\ge-2\)
\(PT\Leftrightarrow x^2-6x-4+\left(x^2+8\right)-3\sqrt{x^3+8}=0\)
\(\Leftrightarrow x^2-6x-4+\frac{\left(x-1\right)\left(x-2\right)\left(x^2-6x+4\right)}{x^2+8+3\sqrt{x^3+8}}=0\)
\(\Leftrightarrow\left(x^2-6x-4\right)\left(1+\frac{\left(x^2-3x+2\right)}{x^2+8+3\sqrt{x^3+8}}\right)=0\)
\(\Leftrightarrow\left(x^2-6x-4\right)\left(\frac{2x^2-3x+10+3\sqrt{x^3+8}}{x^2+8+3\sqrt{x^3+8}}\right)=0\)
Cái ngoặc to hiển nhiên vô nghiệm!
Do đó \(x^2-6x-4=0\)
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ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+2}\ge0\\b=\sqrt{x^2-2x+4}>0\end{matrix}\right.\) \(\Rightarrow b^2-a^2=x^2-3x+2\)
Phương trình trở thành:
\(2\left(b^2-a^2\right)=3ab\)
\(\Leftrightarrow2b^2-3ab-2a^2=0\)
\(\Leftrightarrow\left(b-2a\right)\left(2b+a\right)=0\)
\(\Leftrightarrow b=2a\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)
\(\Leftrightarrow x^2-2x+4=4\left(x+2\right)\)
