Đáp án:
$\begin{array}{l}
1)\\
c)P = \frac{2}{{\sqrt 3 + 1}} - \frac{1}{{\sqrt 3 - 2}} + \frac{{12}}{{\sqrt 3 + 3}}\\
= \frac{2}{{\sqrt 3 + 1}} + \frac{{12}}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} - \frac{{\sqrt 3 + 2}}{{\left( {\sqrt 3 - 2} \right)\left( {\sqrt 3 + 2} \right)}}\\
= \frac{2}{{\sqrt 3 + 1}} + \frac{{4\sqrt 3 }}{{\sqrt 3 + 1}} - \frac{{\sqrt 3 + 2}}{{3 - {2^2}}}\\
= \frac{{2 + 4\sqrt 3 }}{{\sqrt 3 + 1}} + 2 + \sqrt 3 \\
= \frac{{2 + 4\sqrt 3 + \left( {2 + \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 1}}\\
= \frac{{2 + 4\sqrt 3 + 3\sqrt 3 + 5}}{{\sqrt 3 + 1}}\\
= \frac{{7 + 7\sqrt 3 }}{{\sqrt 3 + 1}}\\
= 7\\
2)\\
b)x \ge 0;x \ne 4;x \ne 9\\
B = \frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }} + \frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}\\
= \frac{{\sqrt x + 3}}{{\sqrt x - 2}} - \frac{{\sqrt x + 2}}{{\sqrt x - 3}} + \frac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{x - 9 - \left( {x - 4} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{1}{{\sqrt x - 2}}\\
c)\,x \ge 0;x \ne 4;x \ne 9\\
T = \frac{A}{B} = \left( {1 - \frac{{\sqrt x }}{{1 + \sqrt x }}} \right):\frac{1}{{\sqrt x - 2}}\\
= \frac{1}{{\sqrt x + 1}}.\left( {\sqrt x - 2} \right)\\
= \frac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
= \frac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}}\\
= 1 - \frac{3}{{\sqrt x + 1}}\\
Do\,\sqrt x \ge 0\forall x\,tmdk\\
\Rightarrow \sqrt x + 1 \ge 1\\
\Rightarrow \frac{3}{{\sqrt x + 1}} \le 3\\
\Rightarrow 1 - \frac{3}{{\sqrt x + 1}} \ge - 2
\end{array}$
Vậy GTNN của T = -2 khi và chỉ khi x=0
