Question

giải hệ phương trình ;

\(\hept{\begin{cases}4\left(x+y\right)-5\left(x-y\right)=0\\\frac{40}{x+y}+\frac{40}{x-y}=9\end{cases}}\)

 

Answer 1

\(\hept{\begin{cases}4x+4y-5x+5y=0\\\frac{40\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}+\frac{40\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=\frac{9\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-x+9y=0\\40x-40y+40x+40y=9\left(x^2-y^2\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-x+9=0\\80x=9x^2-9y^2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=9\\9.9^2-9y^2-80.9=0\end{cases}\Rightarrow\hept{\begin{cases}x=9\\-9y^2=9\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\y=1\end{cases}}}\)