Đáp án:
$(x;y)=(16;4)$
Giải thích các bước giải:
ĐKXĐ: $x,y \ge 0;x \ne 0;{x^2} + {y^2} > 0$
c) Ta có:
$M = A + B = \dfrac{2}{{\sqrt x - \sqrt y }}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
M = 1\\
x = 4y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{2}{{\sqrt x - \sqrt y }} = 1\\
x = 4y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt x - \sqrt y = 2\\
\sqrt x = 2\sqrt y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2\sqrt y - \sqrt y = 2\\
\sqrt x = 2\sqrt y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt y = 2\\
\sqrt x = 2\sqrt y
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 4\\
\sqrt x = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 4\\
x = 16
\end{array} \right.
\end{array}$
Vậy $(x;y)=(16;4)$
