TK
x+2y=5 và 3x+4y=5
⇔2x+4y=10 và 3x+4y=5
⇔-x=5 và x+2y=5
⇔x=-5 và -5+2y=5
⇔x=-5 và 2y=10
⇔x=-5 và y=5
\(\left\{{}\begin{matrix}x+2y=5\\3x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+4y=10\\3x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-5+2y=5\\x=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y=10\\x=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-5\end{matrix}\right.\)
\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne-2\\y\ne3\end{matrix}\right.\\ \left\{{}\begin{matrix}\dfrac{3}{x+2}+\dfrac{1}{y-3}=4\\\dfrac{2}{x+2}-\dfrac{3}{y-3}=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+2}+\dfrac{3}{y-3}=12\\\dfrac{2}{x+2}-\dfrac{3}{y-3}=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+2}+\dfrac{1}{y-3}=4\\\dfrac{11}{x+2}=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{1}+\dfrac{1}{y-3}=4\\x+2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3+\dfrac{1}{y-3}=4\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-3}=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-3=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\x=-1\left(tm\right)\end{matrix}\right.\)
