Đáp án:
a) \(\dfrac{{2\left( {\sqrt x - 2} \right)}}{x}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 4\\
A = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {x - 4} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {x - 4} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - x + \sqrt x + 2}}{{\left( {x - 4} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {x - 4} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {x - 4} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {x - 4} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{2\left( {x - 4} \right)}}{{x\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\left( {\sqrt x - 2} \right)}}{x}\\
b)Thay:x = 4 + 2\sqrt 3 = 3 + 2\sqrt 3 .1 + 1 = {\left( {\sqrt 3 + 1} \right)^2}\\
\to A = \dfrac{{2\left( {\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 2} \right)}}{{4 + 2\sqrt 3 }} = \dfrac{{2\sqrt 3 + 2 - 2}}{{4 + 2\sqrt 3 }}\\
= - 3 + 2\sqrt 3 \\
c)A \ge \dfrac{1}{4}\\
\to \dfrac{{2\left( {\sqrt x - 2} \right)}}{x} \ge \dfrac{1}{4}\\
\to \dfrac{{2\sqrt x - 4}}{x} \ge \dfrac{1}{4}\\
\to \dfrac{{8\sqrt x - 16 - x}}{{4x}} \ge 0\\
\to 8\sqrt x - 16 - x \ge 0\left( {do:x > 0} \right)\\
\to - {\left( {\sqrt x - 4} \right)^2} \ge 0\\
\to {\left( {\sqrt x - 4} \right)^2} \le 0\\
\to \sqrt x - 4 = 0\\
\to x = 16
\end{array}\)
