Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x > 0\\
y > 0
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\\
\dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt y }} = 4 - \sqrt x - \sqrt y \\
\Leftrightarrow \left( {\sqrt x + \dfrac{1}{{\sqrt x }} - 2} \right) + \left( {\sqrt y + \dfrac{1}{{\sqrt y }} - 2} \right) = 0\\
\Leftrightarrow \dfrac{{{{\sqrt x }^2} + 1 - 2\sqrt x }}{{\sqrt x }} + \dfrac{{{{\sqrt y }^2} + 1 - 2\sqrt y }}{{\sqrt y }} = 0\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} + \dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y }} = 0\\
\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} \ge 0,\,\,\,\forall x > 0\\
\dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y }} \ge 0,\,\,\,\,\forall y > 0\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} = 0\\
\dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y }} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt x - 1 = 0\\
\sqrt y - 1 = 0
\end{array} \right. \Leftrightarrow x = y = 1\,\,\,\left( {t/m} \right)
\end{array}\)
Em xem lại đề của câu b nhé!!
