Question

2(x²-3x+2)= 3√x³+8

Answer 1

ĐK: \(x\ge-2\)

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

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{x^2-2x+4}=b\end{matrix}\right.\left(a\ge0,b\ge\sqrt{3}\right)\)

\(pt\Leftrightarrow2\left(b^2-a^2\right)=3ab\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=-2b\left(l\right)\\2a=b\end{matrix}\right.\)

\(2a=b\Leftrightarrow2\sqrt{x+2}=\sqrt{x^2-2x+4}\)

\(\Leftrightarrow4x+8=x^2-2x+4\)

\(\Leftrightarrow x^2-6x-4=0\)

\(\Leftrightarrow x=3\pm\sqrt{13}\left(tm\right)\)