Đáp án:
a) 1
b) \(\dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\)
c) \(MaxM = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to A = \dfrac{{3 + 2\sqrt 2 - \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 2}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 3}}\\
= \dfrac{{5 + 2\sqrt 2 - \sqrt 2 - 1}}{{\sqrt 2 + 1 + 3}}\\
= \dfrac{{4 + \sqrt 2 }}{{4 + \sqrt 2 }} = 1\\
b)B = \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 2\left( {\sqrt x + 3} \right) - 3\sqrt x - 4}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 + 2\sqrt x + 6 - 3\sqrt x - 4}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}} = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
c)M = B:A = \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}:\dfrac{{x - \sqrt x + 2}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 1}}{{x - \sqrt x + 2}}\\
\dfrac{1}{M} = \dfrac{{x - \sqrt x + 2}}{{\sqrt x + 1}}\\
= \dfrac{{x + 2\sqrt x + 1 - 3\sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - 3\left( {\sqrt x + 1} \right) + 4}}{{\sqrt x + 1}}\\
= \left( {\sqrt x + 1} \right) - 3 + \dfrac{4}{{\sqrt x + 1}}\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{4}{{\sqrt x + 1}}} = 4\\
\to \left( {\sqrt x + 1} \right) + \dfrac{4}{{\sqrt x + 1}} - 3 \ge 1\\
\to Min\dfrac{1}{M} = 1\\
\to MaxM = 1\\
\Leftrightarrow \left( {\sqrt x + 1} \right) = \dfrac{4}{{\sqrt x + 1}}\\
\to {\left( {\sqrt x + 1} \right)^2} = 4\\
\to \sqrt x + 1 = 2\\
\to x = 1
\end{array}\)
