11: ĐKXĐ: \(x\ne2\)
\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+3\left(y+3\right)=7\\\dfrac{-3}{x-2}+2\left(y+3\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-2}+9\left(y+3\right)=21\\\dfrac{-3}{x-2}+2\left(y+3\right)=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11\left(y+3\right)=22\\\dfrac{1}{x-2}+3\left(y+3\right)=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+3=2\\\dfrac{1}{x-2}=7-3\cdot2=7-6=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\x-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\left(nhận\right)\)
12: ĐKXĐ: \(y\ne2x\)
\(\left\{{}\begin{matrix}\dfrac{1}{2x-y}+x+3y=\dfrac{3}{2}\\\dfrac{4}{2x-y}-5\left(x+3y\right)=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{2x-y}+5\left(x+3y\right)=\dfrac{15}{2}\\\dfrac{4}{2x-y}-5\left(x+3y\right)=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{2x-y}=\dfrac{9}{2}\\\dfrac{1}{2x-y}+\left(x+3y\right)=\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-y=2\\x+3y=\dfrac{3}{2}-\dfrac{1}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=6\\x+3y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=7\\2x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x-2=0\end{matrix}\right.\)(nhận)
11, Đặt \(\dfrac{1}{x-2}=a;y+3=b\)
\(\left\{{}\begin{matrix}a+3b=7\\-3a+2b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7-3b\\-3\left(7-3b\right)+2b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=7-3b\\11b-21=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=1\\y+3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
12, tương tự nhé
