Đáp án: $\begin{array}{l}a)Dkxd:x \ge 0;x \ne 1\\A = \dfrac{{x\sqrt x + 26\sqrt x – 19}}{{x + 2\sqrt x – 3}} – \dfrac{{2\sqrt x }}{{\sqrt x – 1}} + \dfrac{{\sqrt x – 3}}{{\sqrt x + 3}}\\ = \dfrac{{x\sqrt x + 26\sqrt x – 19 – 2\sqrt x \left( {\sqrt x + 3} \right) + \left( {\sqrt x – 3} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{x\sqrt x + 26\sqrt x – 19 – 2x – 6\sqrt x + x – 4\sqrt x + 3}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{x\sqrt x – x + 16\sqrt x – 16}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{x\left( {\sqrt x – 1} \right) + 16\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{\left( {\sqrt x – 1} \right)\left( {x + 16} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{x + 16}}{{\sqrt x + 3}}\\b)P = \dfrac{{x + 16}}{{\sqrt x + 3}} = \dfrac{{x – 9 + 25}}{{\sqrt x + 3}}\\ = \sqrt x – 3 + \dfrac{{25}}{{\sqrt x + 3}}\\ = \sqrt x + 3 + \dfrac{{25}}{{\sqrt x + 3}} – 6\\Do:\sqrt x + 3 > 0\\ \Rightarrow Theo\,Co – si:\\\left( {\sqrt x + 3} \right) + \dfrac{{25}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{25}}{{\sqrt x + 3}}} = 10\\ \Rightarrow A \ge 10 – 6 = 4\\ \Rightarrow A \ge 4\\ \Rightarrow GTNN:A = 4\\Khi:\left( {\sqrt x + 3} \right) = \dfrac{{25}}{{\sqrt x + 3}}\\ \Rightarrow {\left( {\sqrt x + 3} \right)^2} = 25\\ \Rightarrow \sqrt x + 3 = 5\\ \Rightarrow \sqrt x = 2\\ \Rightarrow x = 4\left( {tmdk} \right)\end{array}$ Reply
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{x\sqrt x + 26\sqrt x – 19}}{{x + 2\sqrt x – 3}} – \dfrac{{2\sqrt x }}{{\sqrt x – 1}} + \dfrac{{\sqrt x – 3}}{{\sqrt x + 3}}\\
= \dfrac{{x\sqrt x + 26\sqrt x – 19 – 2\sqrt x \left( {\sqrt x + 3} \right) + \left( {\sqrt x – 3} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x\sqrt x + 26\sqrt x – 19 – 2x – 6\sqrt x + x – 4\sqrt x + 3}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x\sqrt x – x + 16\sqrt x – 16}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x\left( {\sqrt x – 1} \right) + 16\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x – 1} \right)\left( {x + 16} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 16}}{{\sqrt x + 3}}\\
b)P = \dfrac{{x + 16}}{{\sqrt x + 3}} = \dfrac{{x – 9 + 25}}{{\sqrt x + 3}}\\
= \sqrt x – 3 + \dfrac{{25}}{{\sqrt x + 3}}\\
= \sqrt x + 3 + \dfrac{{25}}{{\sqrt x + 3}} – 6\\
Do:\sqrt x + 3 > 0\\
\Rightarrow Theo\,Co – si:\\
\left( {\sqrt x + 3} \right) + \dfrac{{25}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{25}}{{\sqrt x + 3}}} = 10\\
\Rightarrow A \ge 10 – 6 = 4\\
\Rightarrow A \ge 4\\
\Rightarrow GTNN:A = 4\\
Khi:\left( {\sqrt x + 3} \right) = \dfrac{{25}}{{\sqrt x + 3}}\\
\Rightarrow {\left( {\sqrt x + 3} \right)^2} = 25\\
\Rightarrow \sqrt x + 3 = 5\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)
\end{array}$