\(2y^3+y+2x\sqrt{1-x}=3\sqrt{1-x}\)
\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\left(\sqrt{1-x}-y\right)=3\left(\sqrt{1-x}-y\right)\)
\(\Leftrightarrow2y\left(x+y^2-1\right)+2x\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}=3\cdot\dfrac{1-x-y^2}{\sqrt{1-x}+y}\)
\(\Leftrightarrow2y\left(x+y^2-1\right)-2x\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}+3\cdot\dfrac{x+y^2-1}{\sqrt{1-x}+y}=0\)
\(\Leftrightarrow\left(x+y^2-1\right)\left(2y-\dfrac{2x}{\sqrt{1-x}+y}+\dfrac{3}{\sqrt{1-x}+y}\right)=0\)
\(\Leftrightarrow x+y^2-1=0\)\(\Leftrightarrow x=1-y^2\)
Thay vào pt(2) ta có: \(\sqrt{2y^2+1}+y=4+\sqrt{5-y^2}\)
\(\Leftrightarrow y=2\Leftrightarrow x=-3\)
\(2y^3+y+2x\sqrt{1-x}=3\sqrt{1-x}\)
\(\Leftrightarrow2y^3+y=2\sqrt{\left(1-x\right)^3}+\sqrt{1-x}\)
\(\Leftrightarrow y=\sqrt{1-x}\)
Thê vô PT còn lại giải tiêp
