\(a)\) \(\left\{{}\begin{matrix}3x-y=1.\\x+y=3.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x=4.\\x+y=3.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=2.\end{matrix}\right.\)
\(b)\) \(\left\{{}\begin{matrix}3x+2y=5.\\4x+3y=7.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9x+6y=15.\\8x+6y=14.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=1.\end{matrix}\right.\)
\(c)\left\{{}\begin{matrix}\dfrac{1}{x-2}-\dfrac{3}{y+1}=8.\\\dfrac{1}{x-2}+\dfrac{5}{y+1}=16.\end{matrix}\right.\) \(\left(x\ne2;y\ne-1\right).\)
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{y+1}=b\left(a;b\ne0\right).\)
\(\Rightarrow\left\{{}\begin{matrix}a-3b=8.\\a+5b=16.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=11.\\b=1.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=11.\\\dfrac{1}{y+1}=1.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}11x-22=1.\\y+1=1.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{23}{11}.\\y=0.\end{matrix}\right.\) \(\left(TM\right).\)
