Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 + \left( {x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{x + \sqrt x + 1}}\\
b,\\
x = 7 - 4\sqrt 3 = 4 - 2.2.\sqrt 3 + 3 = {\left( {2 - \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = 2 - \sqrt 3 \\
A = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} = \dfrac{{2 - \sqrt 3 }}{{\left( {7 - 4\sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) + 1}} = \dfrac{{2 - \sqrt 3 }}{{10 - 5\sqrt 3 }} = \dfrac{{2 - \sqrt 3 }}{{5.\left( {2 - \sqrt 5 } \right)}} = \dfrac{1}{5}\\
c,\\
A = \dfrac{2}{7} \Leftrightarrow \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} = \dfrac{2}{7}\\
\Leftrightarrow 7\sqrt x = 2x + 2\sqrt x + 2\\
\Leftrightarrow 2x - 5\sqrt x + 2 = 0\\
\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {2\sqrt x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 0\\
2\sqrt x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{4}
\end{array} \right.\\
d,\\
A - \dfrac{1}{3} = \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} - \dfrac{1}{3} = \dfrac{{3\sqrt x - \left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{ - x + 2\sqrt x - 1}}{{x + \sqrt x + 1}} = \dfrac{{ - \left( {x - 2\sqrt x + 1} \right)}}{{x + \sqrt x + 1}} = \dfrac{{ - {{\left( {\sqrt x - 1} \right)}^2}}}{{x + \sqrt x + 1}}\\
x + \sqrt x + 1 \ge 1 > 0,\,\,\,\forall x \ge 0\\
{\left( {\sqrt x - 1} \right)^2} \ge 0,\,\,\,\forall x \ge 0\\
\Rightarrow A - \dfrac{1}{3} \le 0 \Rightarrow A \le \dfrac{1}{3},\,\,\,\forall x \ge 0,x \ne 1\\
\Rightarrow {A_{\max }} = \dfrac{1}{3} \Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 0 \Leftrightarrow x = 1\,\,\,\,\,\left( {L,x \ne 1} \right)
\end{array}\)
Suy ra không tìm được giá trị lớn nhất của A.
