Question

Giải PT:

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

Answer 1

ĐKXĐ: \(x\ge-2\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a>0\\\sqrt{x+2}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2\left(a^2-b^2\right)=3ab\)

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

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

\(\Leftrightarrow a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-2x+4=4\left(x+2\right)\)

\(\Leftrightarrow...\)