Đáp án:
b. \(Min = \dfrac{7}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a \ge 0;x \ne 1\\
b.A = x - \dfrac{{2\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} + \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} + 1\\
= x - 2\sqrt x + \sqrt x + 1 + 1\\
= x - \sqrt x + 2\\
= x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\\
\to Min = \dfrac{7}{4}\\
\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0\\
\Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
