Question

giải/hệ/phương/trình:\(\dfrac{1}{x-2}+\dfrac{5}{2y-1}=3\)

\(\dfrac{3}{x-2}-\dfrac{1}{2y-1}=1\)

Answer 1

`đk:x ne 2,y ne 1/2`

ĐẶt `a=1/(x-2),b=1/(2y-1)`

`hpt<=>` $\begin{cases}a+5b=3\\3a-b=1\\\end{cases}$

`<=>` $\begin{cases}3a+15b=9\\3a-b=1\\\end{cases}$

`<=>` $\begin{cases}16b=8\\a=3-5b\\\end{cases}$

`<=>` $\begin{cases}b=\dfrac12\\a=\dfrac12\\\end{cases}$

`<=>` $\begin{cases}x-2=2\\2y-1=2\\\end{cases}$

`<=>` $\begin{cases}x=4\\y=\dfrac32\\\end{cases}$

Answer 2

Đk: \(x\ne2;y\ne\dfrac{1}{2}\)

Đặt \(a=\dfrac{1}{x-2},b=\dfrac{1}{2y-1}\) (a,b khác 0)

Có hệ: \(\left\{{}\begin{matrix}a+5b=3\\3a-b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\15a-5b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}16a=8\\3a-b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=3a-1=\dfrac{1}{2}\end{matrix}\right.\)(tm)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{1}{2}\\\dfrac{1}{2y-1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3}{2}\end{matrix}\right.\)(tm)