Question

[long ma bac giang]

Answer 1

[HT.Phong (9A5)]

1) Đặt: \(\dfrac{1}{x}=u;\dfrac{1}{y-2}=v\)

\(=>\left\{{}\begin{matrix}2u+3v=4\\4u-v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4u+6v=8\\4u-v=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}7v=7\\4u-v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}v=1\\u=\dfrac{1}{2}\end{matrix}\right.\) 

\(=>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\\\dfrac{1}{y-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\) 

2) Đặt: \(\dfrac{1}{x+1}=u;\dfrac{1}{y}=v\) 

\(=>\left\{{}\begin{matrix}2u+3v=-1\\2u+5v=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u+3v=-1\\2v=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2u=-1\\v=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=-\dfrac{1}{2}\\v=0\end{matrix}\right.\\ =>\left\{{}\begin{matrix}\dfrac{1}{x+1}=-\dfrac{1}{2}\\\dfrac{1}{y}=0\end{matrix}\right.=>x,y\in\varnothing\) 

3) Đặt: \(\dfrac{1}{x}=u;\dfrac{1}{y-2}=v\) 

\(=>\left\{{}\begin{matrix}u-v=-1\\4u+3v=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4u-4v=-4\\4u+3v=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}v=\dfrac{9}{7}\\u=\dfrac{2}{7}\end{matrix}\right.\\ =>\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y-2}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y-2=\dfrac{7}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}+2=\dfrac{11}{2}\end{matrix}\right.\)