Đáp án:
\(\begin{array}{l}
a)\dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}}\\
b)N\left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{4}
\end{array} \right.\\
c)1 > x > 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0\\
N = \dfrac{{x + 2 - x + \sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}.\dfrac{{4\sqrt x }}{3}\\
= \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}}\\
b)N = \dfrac{8}{9}\\
\to \dfrac{{4\sqrt x }}{{3\left( {x - \sqrt x + 1} \right)}} = \dfrac{8}{9}\\
\to \dfrac{{\sqrt x }}{{x - \sqrt x + 1}} = \dfrac{2}{3}\\
\to 3\sqrt x = 2x - 2\sqrt x + 2\\
\to 2x - 5\sqrt x + 2 = 0\\
\to \left( {\sqrt x - 2} \right)\left( {2\sqrt x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 0\\
2\sqrt x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{4}
\end{array} \right.\\
c)\dfrac{1}{N} > \dfrac{{3\sqrt x }}{4}\\
\to \dfrac{{3\left( {x - \sqrt x + 1} \right)}}{{4\sqrt x }} > \dfrac{{3\sqrt x }}{4}\\
\to \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} > \sqrt x \\
\to \dfrac{{x - \sqrt x + 1 - x}}{{\sqrt x }} > 0\\
\to \dfrac{{1 - \sqrt x }}{{\sqrt x }} > 0\\
\to \left\{ \begin{array}{l}
x > 0\\
1 - \sqrt x > 0
\end{array} \right. \to 1 > x > 0
\end{array}\)
