Đáp án:
a. \(A = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0\\
A = 1:\left[ {\dfrac{{x + 2\sqrt x - 2 - \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right]\\
= 1.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{2x + \sqrt x - 1 - x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x + \sqrt x }}\\
= \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
b.Thay:x = 7 - 4\sqrt 3 \\
= 4 - 2.2.\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\to A = \dfrac{{7 - 4\sqrt 3 - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 1}}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}\\
= \dfrac{{8 - 4\sqrt 3 - 2 + \sqrt 3 }}{{2 - \sqrt 3 }}\\
= \dfrac{{6 - 3\sqrt 3 }}{{2 - \sqrt 3 }} = 3\\
c.A = \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x - 1 + \dfrac{1}{{\sqrt x }}\\
= \sqrt x + \dfrac{1}{{\sqrt x }} - 1\\
Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} \\
\to \sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\\
\to \sqrt x + \dfrac{1}{{\sqrt x }} - 1 \ge 2 - 1\\
\to A \ge 1\\
\to MinA = 1\\
\Leftrightarrow \sqrt x = \dfrac{1}{{\sqrt x }}\\
\Leftrightarrow x = 1
\end{array}\)
