Question

Giải pt:

\(2\left(x^2+2\right)=5\sqrt{x^3+1}\)

Answer 1

Answer 2

ĐKXĐ: \(x\ge-1\)

\(2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\) ta được:

\(2\left(a^2+b^2\right)=5ab\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\sqrt{x^2-x+1}\\2\sqrt{x+1}=\sqrt{x^2-x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-5x+3=0\\x^2-5x-3=0\end{matrix}\right.\)