a/ ĐKXĐ: \(x\ge-1\)
\(5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+1}\\b=\sqrt{x^2-x+1}\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)
Phương trình trở thành:
\(5ab=2\left(a^2+b^2\right)\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\sqrt{x^2-x+1}\\2\sqrt{x+1}=\sqrt{x^2-x+1}\end{matrix}\right.\) \(\Leftrightarrow...\)
b/
\(\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\) \(\Rightarrow3x^2-xy-2y^2=0\Leftrightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\y=-\frac{3}{2}x\end{matrix}\right.\)
TH1: \(x=y\Rightarrow2x^2-x^2=1\Rightarrow x^2=1\Rightarrow\left\{{}\begin{matrix}x=...\\y=...\end{matrix}\right.\)
TH2: \(y=-\frac{3}{2}x\Rightarrow2x^2-\frac{9}{4}x^2=1\Rightarrow-\frac{1}{4}x^2=1\) (vô nghiệm)
