\(a,\left\{{}\begin{matrix}\left(x+6y\right)\left(3x-y\right)=0\\x^2-y^2+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-6y\left(1\right)\\y=3x\left(2\right)\end{matrix}\right.\\x^2-y^2+2y=1\left(3\right)\end{matrix}\right.\)
\(\left(1\right)và\left(3\right)\Rightarrow\left(-6y\right)^2-y^2+2y-1=0\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{1}{7}\Rightarrow x=-\dfrac{6}{7}\\y=-\dfrac{1}{5}\Rightarrow x=\dfrac{6}{5}\end{matrix}\right.\)
\(\left(2\right)và\left(3\right)\Rightarrow x^2-\left(3x\right)^2+2.3x=1\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=\dfrac{3}{2}\\x=\dfrac{1}{4}\Rightarrow y=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left\{\left(-\dfrac{6}{7};\dfrac{1}{7}\right);\left(\dfrac{1}{4};\dfrac{3}{4}\right);\left(\dfrac{1}{2};\dfrac{3}{2}\right);\left(\dfrac{6}{5};-\dfrac{1}{5}\right)\right\}\)
b,\(\left\{{}\begin{matrix}x^2-2x-y^2=-1\left(1\right)\\x^2-y^2-4y-4=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-2x+1-y^2=0\Leftrightarrow\left(x-1\right)^2-y^2=0\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y+1\left(3\right)\\x=1-y\left(4\right)\end{matrix}\right.\)
\(thay\left(3\right)vào\left(2\right)\) \(rồi\) \(thay\left(4\right)vào\left(2\right)\Rightarrow\left(x;y\right)=....\)
\(B5:\)
\(a,\left\{{}\begin{matrix}\left(x-y\right)\left(2x+y-1\right)=0\left(1\right)\\\left(x+y\right)\left(2x-y\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=y\left(3\right)\\y=1-2x\left(4\right)\end{matrix}\right.\) \(thay\left(3\right)vào\left(2\right)rồi\) \(thay\left(4\right)vào\left(2\right)\Rightarrow\left(x;y\right)=....\)
\(b,\left\{{}\begin{matrix}\left(x+3y-1\right)\left(2x+y-4\right)=0\left(1\right)\\4x-y=1\Leftrightarrow y=4x-1\left(2\right)\end{matrix}\right.\)
\(thay\left(2\right)vào\left(1\right)\Rightarrow\left(x;y\right)=....\)
