Question

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

Answer 1

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)

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

\(\Rightarrow5ab=2\left(a^2+b^2\right)\)

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

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

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

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