Đáp án:
\(\begin{array}{l}
a)\dfrac{1}{{3 - 2\sqrt x }}\\
b)Max = 3\\
c)\dfrac{{\sqrt 3 - 1}}{{3\sqrt 3 - 7}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
C = \dfrac{{2\sqrt x - 5\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {2\sqrt x - 3} \right)}}:\dfrac{{3 - 3\sqrt x + 2}}{{1 - \sqrt x }}\\
= \dfrac{{ - 3\sqrt x + 5}}{{\left( {\sqrt x - 1} \right)\left( {2\sqrt x - 3} \right)}}.\left( { - \dfrac{{\sqrt x - 1}}{{5 - 3\sqrt x }}} \right)\\
= - \dfrac{1}{{2\sqrt x - 3}}\\
= \dfrac{1}{{3 - 2\sqrt x }}\\
b)C' = \dfrac{1}{C}.\dfrac{1}{{\sqrt x + 1}}\\
= \left( {3 - 2\sqrt x } \right).\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{3 - 2\sqrt x }}{{\sqrt x + 1}} = \dfrac{{ - 2\left( {\sqrt x + 1} \right) + 5}}{{\sqrt x + 1}}\\
= - 2 + \dfrac{5}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{5}{{\sqrt x + 1}} \le 5\\
\to - 2 + \dfrac{5}{{\sqrt x + 1}} \le 3\\
\to Max = 3\\
\Leftrightarrow x = 0\\
c)Thay:x = \dfrac{2}{{2 - \sqrt 3 }}\\
= \dfrac{4}{{4 - 2\sqrt 3 }} = \dfrac{4}{{3 - 2\sqrt 3 .1 + 1}}\\
= \dfrac{4}{{{{\left( {\sqrt 3 - 1} \right)}^2}}}\\
\to C = \dfrac{1}{{3 - 2\sqrt {\dfrac{4}{{{{\left( {\sqrt 3 - 1} \right)}^2}}}} }} = \dfrac{1}{{3 - 2.\dfrac{2}{{\sqrt 3 - 1}}}}\\
= \dfrac{1}{{\dfrac{{3\sqrt 3 - 3 - 4}}{{\sqrt 3 - 1}}}} = \dfrac{{\sqrt 3 - 1}}{{3\sqrt 3 - 7}}
\end{array}\)
