Đáp án + Giải thích các bước giải:
ĐKXĐ : `x \ne0 , y \ne 0`
\(\left\{ \begin{array}{l}\dfrac1x+ \dfrac1y=\dfrac16\\\dfrac4x+\dfrac6y=\dfrac45\end{array} \right.\)
Đặt `a = 1/x` và `b=1/y`
`⇔`\(\left\{ \begin{array}{l}a+b=\dfrac16\\4a+6b=\dfrac45\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}-4a-4b=- \dfrac23\\4a+6b=\dfrac45\end{array} \right.\)
`<=>`\(\left\{ \begin{array}{l}a+b=\dfrac16\\2b=\dfrac2{15}\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}a+b= \dfrac16\\b=\dfrac1{15}\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}a+\dfrac1{15}=\dfrac16\\b=\dfrac1{15}\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}a=\dfrac1{10}\\b=\dfrac1{15}\end{array} \right.\)
`⇔`\(\left\{ \begin{array}{l}\dfrac1x=\dfrac1{10}\\\dfrac1y=\dfrac1{15}\end{array} \right.\)
`<=>`\(\left\{ \begin{array}{l}x=10\\y=15\end{array} \right.\) (TM)
Vậy `(x,y) = (10,15)`
