Đáp án:
\(\begin{array}{l}
1,\\
A = \dfrac{4}{7}\\
2,\\
A + B = \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
x = 16 \Rightarrow A = \dfrac{{\sqrt {16} }}{{\sqrt {16} + 3}} = \dfrac{4}{{4 + 3}} = \dfrac{4}{7}\\
2,\\
A + B = \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{{{\sqrt x }^2} - {3^2}}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x .\left( {\sqrt x - 3} \right) + 2\sqrt x .\left( {\sqrt x + 3} \right) - \left( {3x + 9} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {{{\sqrt x }^2} - 3\sqrt x } \right) + \left( {2{{\sqrt x }^2} + 6\sqrt x } \right) - \left( {3x + 9} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3.\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
