Đáp án:
$\begin{array}{l}
2)B = \dfrac{y}{{{x^2}}} + \dfrac{1}{{xy}}\\
= \dfrac{{{y^2}}}{{{x^2}y}} + \dfrac{x}{{{x^2}y}}\\
= \dfrac{{x + {y^2}}}{{{x^2}y}}\\
C = \left( {\dfrac{x}{{{x^2} - 64}} - \dfrac{{x - 8}}{{{x^2} + 8x}}} \right):\dfrac{{2x - 6}}{{{x^2} + 8x}}\\
+ \dfrac{x}{{8 - x}}\\
= \left[ {\dfrac{x}{{\left( {x - 8} \right)\left( {x + 8} \right)}} - \dfrac{{x - 8}}{{x\left( {x + 8} \right)}}} \right]\\
.\dfrac{{x\left( {x + 8} \right)}}{{2\left( {x - 3} \right)}} - \dfrac{x}{{x - 8}}\\
= \dfrac{{{x^2} - {{\left( {x - 8} \right)}^2}}}{{x\left( {x + 8} \right)\left( {x - 8} \right)}}.\dfrac{{x\left( {x + 8} \right)}}{{2\left( {x - 3} \right)}} - \dfrac{x}{{x - 8}}\\
= \dfrac{{{x^2} - {x^2} + 16x - 64}}{{x - 8}}.\dfrac{1}{{2\left( {x - 3} \right)}} - \dfrac{x}{{x - 8}}\\
= \dfrac{{8x - 32}}{{x - 8}}.\dfrac{1}{{x - 3}} - \dfrac{x}{{x - 8}}\\
= \dfrac{{8x - 32 - x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x - 8} \right)}}\\
= \dfrac{{8x - 32 - {x^2} + 3x}}{{\left( {x - 3} \right)\left( {x - 8} \right)}}\\
= \dfrac{{ - {x^2} + 11x - 32}}{{\left( {x - 3} \right)\left( {x - 8} \right)}}\\
3)A = \dfrac{{2{x^2} - 16x + 43}}{{{x^2} - 8x + 22}}\\
= \dfrac{{2\left( {{x^2} - 8x + 22} \right) - 44 + 43}}{{{x^2} - 8x + 22}}\\
= 2 - \dfrac{1}{{{x^2} - 8x + 22}}\\
Do:{x^2} - 8x + 22\\
= {x^2} - 2.x.4 + 16 + 6\\
= {\left( {x - 4} \right)^2} + 6 \ge 6\\
\Rightarrow \dfrac{1}{{{x^2} - 8x + 22}} \le \dfrac{1}{6}\\
\Rightarrow - \dfrac{1}{{{x^2} - 8x + 22}} \ge - \dfrac{1}{6}\\
\Rightarrow 2 - \dfrac{1}{{{x^2} - 8x + 22}} \ge 2 - \dfrac{1}{6} = \dfrac{{11}}{6}\\
\Rightarrow A \ge \dfrac{{11}}{6}\\
\Rightarrow GTNN:A = \dfrac{{11}}{6}\\
Khi:x = 4
\end{array}$
