Ta có: \(\hept{\begin{cases}xy+x-2=0\\x^2\left(2x-y\right)+y^2=x\left(2y-x\right)+y\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y+1=\frac{2}{x}\\2x^3-x^2y+y^2-2xy+x^2-y=0\end{cases}}\Rightarrow\hept{\begin{cases}y+1=\frac{2}{x}\\\left(2x^3-2xy\right)-\left(x^2y-y^2\right)+\left(x^2-y\right)=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y+1=\frac{2}{x}\\2x\left(x^2-y\right)-y\left(x^2-y\right)+\left(x^2-y\right)=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}y+1=\frac{2}{x}\left(1\right)\\\left(2x-y+1\right)\left(x^2-y\right)=0\left(2\right)\end{cases}}\).
Có (2x-y+1)(x\(^2\)-y)=0
\(\Rightarrow2x-y+1=0\)hoặc \(x^2-y=0\)
\(\Rightarrow y=2x+1\)hoặc \(x^2=y\)
Xét y = 2x + 1 từ (1) \(\Rightarrow2x+1+1=\frac{2}{x}\)
\(\Rightarrow2x+2=\frac{2}{x}\)\(\Rightarrow2x^2+2x-2=0\)\(\Rightarrow2\left(x^2+x-1\right)=0\)
\(\Rightarrow x^2+x-1=0\Rightarrow x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{5}{4}=0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Rightarrow x+\frac{1}{2}=\frac{\sqrt{5}}{2}\)hoặc \(x+\frac{1}{2}=\frac{-\sqrt{5}}{2}\)
\(\Rightarrow x=\frac{\sqrt{5}-1}{2}\)hoặc \(x=\frac{-\sqrt{5}-1}{2}\)
\(\Rightarrow y+1=\frac{2}{\frac{\sqrt{5}-1}{2}}\)hoặc \(y+1=\frac{2}{\frac{-\sqrt{5}-1}{2}}\)
\(\Rightarrow y=\sqrt{5}\)hoặc \(y=-\sqrt{5}\).
Xét \(x^2=y\)từ (1), ta có: \(x^2+1=\frac{2}{x}\)
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Chúc bn hc tốt!
\(PT\Leftrightarrow\hept{\begin{cases}y=\frac{2}{x+1}\\2x^3-x^2y+x^2+y^2-2xy-y=0\end{cases}}\)
Thay y theo x vào phương trình 2 ta được:
\(2x^3-x^3.\frac{2}{x+1}+x^2+\frac{4}{\left(x+1\right)^2}-2x.\frac{2}{x+1}-\frac{2}{x+1}=0\)
\(\Leftrightarrow2x^3\left(x+1\right)^2-2x^3\left(x+1\right)+x^2\left(x+1\right)+4-4x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow2x^5+5x^4+2x^3-5x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2+3x-1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow x=1;x=\frac{1}{2}\left(-3\pm\sqrt{17}\right)\)
Đến đây dễ rồi..
