Đáp án:
\(\begin{array}{l}
a)A = \dfrac{2}{3}\\
b)B = \dfrac{1}{{\sqrt x - 2}}\\
c)0 \le x < 4\\
d)Min = 2\sqrt 3 - 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = \dfrac{1}{4}\\
\to A = \dfrac{1}{{\sqrt {\dfrac{1}{4}} + 1}} = \dfrac{1}{{\dfrac{1}{2} + 1}} = \dfrac{2}{3}\\
b)B = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 9 - x + 4 + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{1}{{\sqrt x - 2}}\\
c)M = A:B = \dfrac{1}{{\sqrt x + 1}}:\dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
M < 0\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 4\\
d)N = \sqrt x - M\\
= \sqrt x - \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{x + \sqrt x - \sqrt x + 2}}{{\sqrt x + 1}}\\
= \dfrac{{x + 2}}{{\sqrt x + 1}} = \dfrac{{x - 1 + 3}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 3}}{{\sqrt x + 1}}\\
= \left( {\sqrt x - 1} \right) + \dfrac{3}{{\sqrt x + 1}}\\
= \left( {\sqrt x + 1} \right) + \dfrac{3}{{\sqrt x + 1}} - 2\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 1} \right) + \dfrac{3}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{3}{{\sqrt x + 1}}} = 2\sqrt 3 \\
\to \left( {\sqrt x + 1} \right) + \dfrac{3}{{\sqrt x + 1}} - 2 \ge 2\sqrt 3 - 2\\
\to Min = 2\sqrt 3 - 2\\
\Leftrightarrow \left( {\sqrt x + 1} \right) = \dfrac{3}{{\sqrt x + 1}}\\
\to {\left( {\sqrt x + 1} \right)^2} = 3\\
\to \sqrt x + 1 = \sqrt 3 \\
\to \sqrt x = \sqrt 3 - 1\\
\to x = 4 - 2\sqrt 3
\end{array}\)
