a: \(\left\{{}\begin{matrix}2x-3y=1\\-x+4y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3y=1\\-2x+8y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=15\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\2x=3y+1=3\cdot3+1=10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\sqrt{2}x+y=1+\sqrt{2}\\x+\sqrt{2}y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\sqrt{2}+y=1+\sqrt{2}\\x\sqrt{2}+2y=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\sqrt{2}+y-x\sqrt{2}-2y=1+\sqrt{2}+\sqrt{2}\\x+\sqrt{2}y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-y=1+2\sqrt{2}\\x=-1-y\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-2\sqrt{2}-1\\x=-1-\sqrt{2}\left(-2\sqrt{2}-1\right)=-1+4+\sqrt{2}=3+\sqrt{2}\end{matrix}\right.\)
c: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{2}{x}+\dfrac{3}{y}=\dfrac{2}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=\dfrac{1}{3}\\\dfrac{2}{x}+\dfrac{3}{y}=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{y}=\dfrac{1}{3}-\dfrac{2}{5}=\dfrac{-1}{15}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=15\\\dfrac{1}{x}=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{5-2}{30}=\dfrac{3}{30}=\dfrac{1}{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=15\\x=10\end{matrix}\right.\left(nhận\right)\)
d: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy-2x+5y-10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+3y=8\\3x-y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-3x+9y=24\\3x-y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8y=40\\x-3y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=3y-8=3\cdot5-8=7\end{matrix}\right.\)
