Đáp án: a) \(\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\) Giải thích các bước giải: \(\begin{array}{l}a)A = \dfrac{{\sqrt x }}{{\sqrt x – 1}} + \dfrac{1}{{\sqrt x + 2}} – \dfrac{{3\sqrt x }}{{x + \sqrt x – 2}}\\ = \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x – 1 – 3\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{x + 2\sqrt x – 2\sqrt x – 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{x – 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\b)S = A.B = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}.\dfrac{{\sqrt x + 3}}{{\sqrt x + 1}}\\ = \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 1}}{{\sqrt x + 2}} = 1 + \dfrac{1}{{\sqrt x + 2}}\\Do:\sqrt x \ge 0\forall x \ge 0;\\ \to \sqrt x + 2 \ge 2\\ \to \dfrac{1}{{\sqrt x + 2}} \le \dfrac{1}{2}\\ \to 1 + \dfrac{1}{{\sqrt x + 2}} \le \dfrac{3}{2}\\ \to MaxS = \dfrac{3}{2}\\ \Leftrightarrow x = 0\end{array}\) Reply
Đáp án:
a) \(\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\sqrt x }}{{\sqrt x – 1}} + \dfrac{1}{{\sqrt x + 2}} – \dfrac{{3\sqrt x }}{{x + \sqrt x – 2}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) + \sqrt x – 1 – 3\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{x + 2\sqrt x – 2\sqrt x – 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{x – 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
b)S = A.B = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}.\dfrac{{\sqrt x + 3}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 1}}{{\sqrt x + 2}} = 1 + \dfrac{1}{{\sqrt x + 2}}\\
Do:\sqrt x \ge 0\forall x \ge 0;\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{1}{{\sqrt x + 2}} \le \dfrac{1}{2}\\
\to 1 + \dfrac{1}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to MaxS = \dfrac{3}{2}\\
\Leftrightarrow x = 0
\end{array}\)