Đáp án + giải thích các bước giải:
$ \left\{\begin{matrix} x+\dfrac{x+y}{y}=\dfrac{1}{2}(1)\\y+\dfrac{x+y}{y}=\dfrac{5}{2}(2) \end{matrix}\right.$
Lấy (2)`-`(1) vế theo vế
`->y+(x+y)/y-x-(x+y)/y=5/2-1/2`
`->y-x=2`
`->y=x+2`(3)
Thế (3) vào (2), có:
`x+2+(x+x+2)/(x+2)=5/2`
`->x+2+(2x+2)/(x+2)=5/2`
`->2(x+2)^2+2(2x+2)=5(x+2)`
`->2(x^2+4x+4)+4x+4=5x+10`
`->2x^2+8x+8+4x+4=5x+10`
`->2x^2+7x+2=0`
`->4x^2+14x+4=0`
`->(2x)^2+2.2x. 7/2+49/4-33/4=0`
`->(2x+7/2)=33/4`
`->`\(\left[ \begin{array}{l}2x+\dfrac{7}{2}=\dfrac{\sqrt{33}}{2}\\2x+\dfrac{7}{2}=-\dfrac{\sqrt{33}}{2}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{33}-7}{4}\\x=\dfrac{-\sqrt{33}-7}{4}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}y=\dfrac{\sqrt{33}-7}{4}+2=\dfrac{\sqrt{33}+1}{4}\\y=\dfrac{-\sqrt{33}-7}{4}+2=\dfrac{-\sqrt{33}+1}{4}\end{array} \right.\)
Vậy `(x;y)=(\frac{\sqrt{33}-7}{4};\frac{\sqrt{33}+1}{4});(\frac{-\sqrt{33}-7}{4};\frac{-\sqrt{33}+1}{4})`
