\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)

\(\Rightarrow x\left(1-2y\right)=5.8\Rightarrow x\left(1-2y\right)=40\)

... ( tự làm tiếp )

\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}=\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}=\dfrac{3}{4}\)

Vì : \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)

Vậy x = -1 ; y = 3