## giúp em vs ạ em cần gấp lắm

Question

giúp em vs ạ em cần gấp lắm

in progress 0
9 months 2020-10-28T18:16:26+00:00 1 Answers 57 views 0

d. $$Max = \dfrac{2}{3}$$
$$\begin{array}{l} A = \dfrac{{15\sqrt x – 11 – \left( {3\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) – \left( {2\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{15\sqrt x – 11 – 3x – 7\sqrt x + 6 – 2x – \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{ – 5x + 7\sqrt x – 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\left( {2 – 5\sqrt x } \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{2 – 5\sqrt x }}{{\sqrt x + 3}}\\ b.A = \dfrac{1}{2}\\ \to \dfrac{{2 – 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\ \to 4 – 10\sqrt x = \sqrt x + 3\\ \to 11\sqrt x = 1\\ \to \sqrt x = \dfrac{1}{{11}}\\ \to x = \dfrac{1}{{121}}\\ c.Thay:x = 3 – 2\sqrt 2 \\ = 2 – 2\sqrt 2 .1 + 1 = {\left( {\sqrt 2 – 1} \right)^2}\\ \to A = \dfrac{{2 – 5\sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} + 3}}\\ = \dfrac{{2 – 5\left( {\sqrt 2 – 1} \right)}}{{\sqrt 2 – 1 + 3}} = \dfrac{{7 – 5\sqrt 2 }}{{\sqrt 2 – 2}}\\ d.A = \dfrac{{2 – 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{{ – 5\left( {\sqrt x + 3} \right) + 17}}{{\sqrt x + 3}}\\ = – 5 + \dfrac{{17}}{{\sqrt x + 3}}\\ Do:\sqrt x \ge 0 \to \sqrt x + 3 \ge 3\\ \to \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{{17}}{3}\\ \to – 5 + \dfrac{{17}}{{\sqrt x + 3}} \le – 5 + \dfrac{{17}}{3}\\ \to – 5 + \dfrac{{17}}{{\sqrt x + 3}} \le \dfrac{2}{3}\\ \to Max = \dfrac{2}{3}\\ \Leftrightarrow x = 0 \end{array}$$