Đáp án:
\(A = \dfrac{{23}}{{\sqrt x + 11}}\)
Giải thích các bước giải:
\(\begin{array}{l}
b.B = \dfrac{{\sqrt {x + 1} + \sqrt x }}{{x + 1 - x}} - \dfrac{{\sqrt {x + 1} - \sqrt x }}{{x + 1 - x}}\\
= \dfrac{{\sqrt {x + 1} + \sqrt x - \sqrt {x + 1} + \sqrt x }}{1}\\
= 2\sqrt x \\
b.A = \left[ {\dfrac{{2\sqrt x + x - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x + 11}}{{23\left( {x + \sqrt x + 1} \right)}}} \right]\\
= \dfrac{{2\sqrt x + x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{23\left( {x + \sqrt x + 1} \right)}}{{\sqrt x + 11}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{23\left( {x + \sqrt x + 1} \right)}}{{\sqrt x + 11}}\\
= \dfrac{{23}}{{\sqrt x + 11}}
\end{array}\)
