cho A=(x+√x+10/x-9 - 1/√x-3）/1/√x-3 a) rút gọn A b)tính A khi x=4 c)Tìm gtnn của A(cô si)

Đáp án:

$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 9\\
A = \left( {\dfrac{{x + \sqrt x  + 10}}{{x - 9}} - \dfrac{1}{{\sqrt x  - 3}}} \right):\dfrac{1}{{\sqrt x  - 3}}\\
 = \dfrac{{x + \sqrt x  + 10 - \left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}.\left( {\sqrt x  - 3} \right)\\
 = \dfrac{{x + \sqrt x  + 10 - \sqrt x  - 3}}{{\sqrt x  + 3}}\\
 = \dfrac{{x + 7}}{{\sqrt x  + 3}}\\
b)x = 4\left( {tmdk} \right)\\
 \Leftrightarrow \sqrt x  = 2\\
 \Leftrightarrow A = \dfrac{{4 + 7}}{{2 + 3}} = \dfrac{{11}}{5}\\
c)A = \dfrac{{x + 7}}{{\sqrt x  + 3}}\\
 = \dfrac{{x - 9 + 16}}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) + 16}}{{\sqrt x  + 3}}\\
 = \sqrt x  - 3 + \dfrac{{16}}{{\sqrt x  + 3}}\\
 = \sqrt x  + 3 + \dfrac{{16}}{{\sqrt x  + 3}} - 6\\
 \ge 2\sqrt {\left( {\sqrt x  + 3} \right).\dfrac{{16}}{{\sqrt x  + 3}}}  - 6\\
 \Leftrightarrow A \ge 2\sqrt {16}  - 6\\
 \Leftrightarrow A \ge 2\\
 \Leftrightarrow GTNN:A = 2\\
Khi:\left( {\sqrt x  + 3} \right) = \dfrac{{16}}{{\sqrt x  + 3}}\\
 \Leftrightarrow \sqrt x  + 3 = 4\\
 \Leftrightarrow x = 1\left( {tmdk} \right)
\end{array}$