Đáp án:
b) \(x = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{7}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{7}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{7}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 1}}{{x + \sqrt x + 1}}.\dfrac{7}{{\sqrt x - 1}}\\
= \dfrac{7}{{x + \sqrt x + 1}}\\
b)P = 4\\
\to \dfrac{7}{{x + \sqrt x + 1}} = 4\\
\to 7 = 4x + 4\sqrt x + 4\\
\to 4x + 4\sqrt x - 3 = 0\\
\to \left( {2\sqrt x - 1} \right)\left( {2\sqrt x + 3} \right) = 0\\
\to 2\sqrt x - 1 = 0\left( {do:2\sqrt x + 3 > 0\forall x} \right)\\
\to x = \dfrac{1}{4}
\end{array}\)
