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