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a) \(\left\{{}\begin{matrix}x^2+3y=1\\3x^2-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+3y=1\\9x^2-3y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x^2=4\\y=3x^2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{\sqrt{10}}{5}\\y=\dfrac{1}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-\sqrt{10}}{5}\\y=\dfrac{1}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\left(x+1\right)\left(y+1\right)=8\\ \Rightarrow xy+x+y+1=8\\ \Rightarrow xy+x+y=7\)
\(x\left(x+1\right)+y\left(y+1\right)+xy=17\\ \Rightarrow x^2+y^2+x+y+xy=17\\ \Rightarrow x^2+y^2=10\)
a)\(\hept{\begin{cases}2x-3y=1\\4x-5y=2\end{cases}\Leftrightarrow\hept{\begin{cases}4x-6y=2\\4x-5y=2\end{cases}}}\)
Trừ 2 vế lại ta được
\(4x-4x-6y+5y=0\Leftrightarrow-y=0\Leftrightarrow y=0\)
\(\Rightarrow x=\frac{1}{2}\)
b)Đặt $S=x+y,P=xy$ thì được:
\(\left\{ \begin{align} & S+P=2+3\sqrt{2} \\ & {{S}^{2}}-2P=6 \\ \end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}\)
\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} S = 2 + \sqrt 2 \\ P = 2\sqrt 2 \end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\\ \left\{ \begin{array}{l} S = - 4 - \sqrt 2 \\ P = 6 + 4\sqrt 2 \end{array} \right.\left( {VN} \right) \end{array} \)
\( c)\left\{ \begin{array}{l} 2{x^2} + xy + 3{y^2} - 2y - 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x + y - 2} \right)^2} = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + y - 2 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 1 \end{array} \right. \)
