Đáp án:
1) 1
2) \(\dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\)
3) \(Min = 2\sqrt 2 + 1\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
A = \dfrac{{3 + 2\sqrt 2 - \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 2}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 3}}\\
= \dfrac{{2\sqrt 2 + 5 - \sqrt 2 - 1}}{{\sqrt 2 + 1 + 3}}\\
= \dfrac{{\sqrt 2 + 4}}{{\sqrt 2 + 4}} = 1\\
2)B = \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 2\left( {\sqrt x + 3} \right) - 5\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 + 2\sqrt x + 6 - 5\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\\
3)M = A:B = \dfrac{{x - \sqrt x + 2}}{{\sqrt x + 3}}:\dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\\
= \dfrac{{x - \sqrt x + 2}}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1 + \sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2} + \left( {\sqrt x - 1} \right) + 2}}{{\sqrt x - 1}}\\
= \left( {\sqrt x - 1} \right) + 1 + \dfrac{2}{{\sqrt x - 1}}\\
Do:x > 1\\
BDT:Co - si:\left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} \ge 2\sqrt {\left( {\sqrt x - 1} \right).\dfrac{2}{{\sqrt x - 1}}} = 2\sqrt 2 \\
\to \left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} \ge 2\sqrt 2 + 1\\
\to Min = 2\sqrt 2 + 1\\
\Leftrightarrow \left( {\sqrt x - 1} \right) = \dfrac{2}{{\sqrt x - 1}}\\
\to {\left( {\sqrt x - 1} \right)^2} = 2\\
\to \sqrt x - 1 = \sqrt 2 \\
\to \sqrt x = 1 + \sqrt 2 \\
\to x = 3 + 2\sqrt 2
\end{array}\)
