Question

Giải hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{96}{x+y}+\dfrac{96}{x-y}=14\\\dfrac{96}{x-y}+\dfrac{72}{x-y}=\dfrac{24}{y}\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x\ne_-^+y;y\ne0\)

Từ PT thứ 2 ta có:\(\dfrac{96}{x-y}+\dfrac{72}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{168}{7y}\)

<=>x-y=7y

<=>x=8y

Thay x=8y vào PT thứ nhất:

\(\dfrac{96}{8y+y}+\dfrac{96}{8y-y}=14\)

<=>\(\dfrac{32}{3y}+\dfrac{96}{7y}=14\)

<=>32.7y+96.3y=294y²

<=>512y=294y²

<=>y=\(\dfrac{256}{147}\left(Doy\ne0\right)\)

=>x=8y=\(\dfrac{2048}{147}\)

Vậy...