Đáp án:
b) \(B = \dfrac{{2x + \sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 909\\
\to A = \dfrac{{\sqrt {909} }}{{\sqrt {909} - 1}} = \dfrac{{3\sqrt {101} }}{{3\sqrt {101} - 1}}\\
b)B = \dfrac{{3x}}{{{{\left( {\sqrt x - 1} \right)}^2}}} - \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{3x - \sqrt x \left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{3x - x + \sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}} = \dfrac{{2x + \sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
c)P = B:A = \dfrac{{2x + \sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {2\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}.\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x - 1}}
\end{array}\)
