\(PT\left(1\right)\Leftrightarrow\left(x^2+y^2\right)\left(x+y\right)+2xy=x+y\\ \Leftrightarrow\left[\left(x+y\right)^2-2xy\right]\left(x+y\right)+2xy-\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)^3-2xy\left(x+y\right)+2xy-\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-1\right]-2xy\left(x+y-1\right)=0\\ \Leftrightarrow\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)-2xy\left(x+y-1\right)=0\\ \Leftrightarrow\left(x+y-1\right)\left[\left(x+y\right)\left(x+y+1\right)-2xy\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y-1=0\\x^2+2xy+x+y^2+y+1=0\left(3\right)\end{matrix}\right.\\ \left(3\right)\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \Leftrightarrow\left(x+y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.n_o\right)\)
Từ đó em thế vô PT(2) thôi
Thế \(x+y-1=0\Leftrightarrow y=x-1\) vào PT(2)
\(\Leftrightarrow\sqrt{x+x-1}=x^2-x+1\\ \Leftrightarrow\sqrt{2x-1}=x^2-x+1\left(x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-1=x^2-x\\ \Leftrightarrow\dfrac{2x-2}{\sqrt{2x-1}+1}-x\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{2}{\sqrt{2x-1}+1}-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\sqrt{2x-1}+x=2\left(4\right)\end{matrix}\right.\)
Giải (4) ta được \(x=1\Leftrightarrow y=0\)
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