Đáp án:
M=$\frac{-88}{7}$
Giải thích các bước giải:
\(
\begin{array}{l}
\left\{ {\begin{array}{*{20}c}
{\frac{1}{x} + \frac{1}{y} = \frac{{ - 5}}{6}} \\
{\frac{{x + 1}}{x} + \frac{3}{y} = \frac{{ - 1}}{3}} \\
\end{array}} \right. \\
Dk:x;y \ne 0 \\
\Leftrightarrow \left\{ {\begin{array}{*{20}c}
{\frac{1}{x} + \frac{1}{y} = \frac{{ - 5}}{6}} \\
{1 + \frac{1}{x} + \frac{3}{y} = \frac{{ - 1}}{3}} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{\frac{1}{x} + \frac{1}{y} = \frac{{ - 5}}{6}} \\
{\frac{1}{x} + \frac{3}{y} = \frac{{ - 4}}{3}} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{\frac{1}{x} = \frac{{ - 7}}{{12}}} \\
{\frac{1}{y} = \frac{{ - 1}}{4}} \\
\end{array}} \right. \\
\Leftrightarrow \left\{ {\begin{array}{*{20}c}
{x = \frac{{ - 12}}{7}} \\
{y = - 4} \\
\end{array}} \right.(tm) \\
\end{array}
\)
=> M=$x_{0}$+ $y_{0}$- $x_{0}$$y_{0}$ =$\frac{-12}{7}$+(-4)-( $\frac{-12}{7}$)(-4)= $\frac{-88}{7}$
