Question

2(x²+2)=5\(\sqrt{x^3+1}\)

Answer 1

ĐKXĐ: \(x\ge-1\)

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

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

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

\(\Leftrightarrow2a^2-5ab+2b^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)=x^2-x+1\\x+1=4\left(x^2-x+1\right)\end{matrix}\right.\)