Question

Giải PT:\(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)

Answer 1

ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{x^2-2x+4}=b\end{matrix}\right.\) pt trở thành:

\(2\left(a^2-b^2\right)=3ab\)

\(\Leftrightarrow2a^2-3ab-2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+4=4x^2-8x+16\\4x+8=x^2-2x+4\end{matrix}\right.\) \(\Rightarrow...\)