1: \(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=8\\2x-3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y-2x+3y=8-1\\x+2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7y=7\\x+2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\x=4-2\cdot1=2\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2x+3y=-5\\6x-5y=27\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6x+9y=-15\\6x-5y=27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14y=-42\\2x+3y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3\\2x=-5-3y=-5-3\cdot\left(-3\right)=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
3: \(\left\{{}\begin{matrix}x+7y=16\\x-3y=-24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+7y-x+3y=16-\left(-24\right)\\x+7y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10y=40\\x+7y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\\x=16-7\cdot4=-12\end{matrix}\right.\)
4: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{1}{y}=1+1:\dfrac{7}{2}=\dfrac{9}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)(nhận)
5: \(\left\{{}\begin{matrix}x\sqrt{2}+y=\dfrac{9}{2}\\3\sqrt{2}x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x\sqrt{2}+2y=9\\3\sqrt{2}x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{2}\cdot x=14\\x\sqrt{2}+y=4,5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{14}{5\sqrt{2}}=\dfrac{7\sqrt{2}}{5}\\y=4,5-x\sqrt{2}=4,5-\dfrac{14}{5}=1,7\end{matrix}\right.\)
6: \(\left\{{}\begin{matrix}x+y\sqrt{5}=0\\x\sqrt{5}+3y=1-\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\sqrt{5}+5y=0\\x\sqrt{5}+3y=1-\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\sqrt{5}+5y-x\sqrt{5}-3y=0-1+\sqrt{5}\\x=-y\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=\sqrt{5}-1\\x=-y\sqrt{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{5}-1}{2}\\x=-\sqrt{5}\cdot\dfrac{\sqrt{5}-1}{2}=\dfrac{-5+\sqrt{5}}{2}\end{matrix}\right.\)
7: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{13}{36}\\\dfrac{6}{x}+\dfrac{10}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{12}{x}+\dfrac{9}{y}=\dfrac{13}{12}\\\dfrac{12}{x}+\dfrac{20}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y}=\dfrac{13}{12}-2=\dfrac{-11}{12}\\\dfrac{6}{x}+\dfrac{10}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=12\\\dfrac{6}{x}=1-\dfrac{10}{12}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=12\\x=36\end{matrix}\right.\left(nhận\right)\)
8: ĐKXĐ: x<>2 và y<>1
\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=6\\x-2=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{13}{4}\end{matrix}\right.\left(nhận\right)\)
9: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{2}{y}=1\\\dfrac{2}{x}+\dfrac{4}{y}=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{2}{y}=1\\\dfrac{1}{x}+\dfrac{2}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}=4\\\dfrac{3}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\\dfrac{2}{y}=\dfrac{3}{1}-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(nhận\right)\)
10: \(\left\{{}\begin{matrix}x\sqrt{2}-y\sqrt{3}=1\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\sqrt{2}-y\sqrt{3}+x+y\sqrt{3}=1+\sqrt{2}\\x+\sqrt{3}y=\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(\sqrt{2}+1\right)=\sqrt{2}+1\\\sqrt{3}y=\sqrt{2}-x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y\sqrt{3}=\sqrt{2}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{\sqrt{2}-1}{\sqrt{3}}\end{matrix}\right.\)
