ĐKXĐ : `x \ne 0 , y \ne 0`
\(\left\{ \begin{array}{l}\dfrac1x+\dfrac1y=\dfrac12\\x+3=y\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}\dfrac1x+\dfrac1{x+3}=\dfrac12\\x+3=y\end{array} \right.\)
Giải pt : `1/x + 1/(x+3) = 1/2`
`⇔ 1/x+1/(x+3) - 1/2 = 0`
`⇔ (2(x+3)+2x-x(x-3))/(2x(x+3)) = 0`
`⇔ (2x+6+2x-x^2-3x)/(2x(x+3)) = 0`
`⇔ (x^2-x-6)/(2x(x+3)) = 0`
`⇔ x^2 - x - 6 = 0`
`⇔ x^2 + 2x - 3x - 6 = 0`
`⇔ x(x+2)-3(x+2) = 0`
`⇔ (x+2)(x-3) = 0`
`1)` `x + 2 = 0 ⇔ x = -2`
`2)` `x - 3 = 0 ⇔ x = 3`
`⇔`\(\left\{ \begin{array}{l}\left[\begin{matrix}x=-2\\x=3\end{matrix}\right.\\\left[\begin{matrix} -2+3=y\\3+3=y\end{matrix}\right.\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}\left[\begin{matrix}x=-2\\x=3\end{matrix}\right.\\\left[\begin{matrix} y=1\\y=6\end{matrix}\right.\end{array} \right.\) (TM)
Vậy `(x,y) = {(-2,1);(3,6)}`
