ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2+x+1\right)}=\frac{2\left(x^2+2\right)}{5}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)
Phương trình trở thành:
\(ab=\frac{2\left(a^2+b^2\right)}{5}\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\sqrt{x^2-x+1}\\2\sqrt{x+1}=\sqrt{x^2-x+1}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+1=4\left(x^2-x+1\right)\\4\left(x+1\right)=x^2-x+1\end{matrix}\right.\) \(\Leftrightarrow...\)
mình bị hack nick mọi người nhé
