Đặt
\(y=x^2-1\)
\(\Rightarrow x^2=y-1\) và
\(x=\sqrt{y}-1\)
Phương trình tương đương
\(y^2+3y\left(\sqrt{y}-1\right)+2\left(\sqrt{y}-1\right)^2\)
\(=y^2+2\cdot\dfrac{3}{2}\cdot y\cdot\left(\sqrt{y}-1\right)+\dfrac{9}{4}\left(\sqrt{y}-1\right)^2=\dfrac{1}{4}\left(\sqrt{y}-1\right)^2\)
\(\Rightarrow\left(y+\dfrac{3}{2}\left(\sqrt{y}-1\right)\right)^2=\left[\dfrac{1}{2}\cdot\left(\sqrt{y}-1\right)\right]^2\)
\(\Rightarrow y+\dfrac{3}{2}\left(\sqrt{y}-1\right)=\dfrac{1}{2}\cdot\left(\sqrt{y}-1\right)\)
\(\Rightarrow y+\sqrt{y}+1=0\)
Mà \(y+\dfrac{1}{2}\cdot2\cdot\sqrt{y}+\dfrac{1}{4}+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
=> Không có y thỏa mãn
\(\Rightarrow PTVN\)
