Question

[Nguyễn Đại Nghĩa]

 gíugiúp vs àaaaaaaaaa

Answer 1

[HT.Phong (9A5)]

a) 

\(\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\left(x\ne1;x\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{2}{y}=-2\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{5}{x-1}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{1}{y}=-1\\x-1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-1-1=-2\\x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=2\end{matrix}\right.\left(tm\right)\)

b) 

\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\left(x\ne2;y\ne-1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2}+\dfrac{2}{y+1}=6\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y+1}=5\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=1\\\dfrac{2}{x-2}+1=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\\dfrac{2}{x-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x-2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\left(tm\right)\)

Answer 2

[HT.Phong (9A5)]

c)

\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\left(x\ne2;y\ne1\right) \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=4\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{5}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{3}{5}=2\\y-1=\dfrac{5}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=2-\dfrac{3}{5}=\dfrac{7}{5}\\y=\dfrac{5}{3}+1=\dfrac{8}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+2=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\left(tm\right)\)