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