Đáp án:
$\begin{array}{l}
B3)\\
a)x = 6\left( {tmdk} \right)\\
\Rightarrow B = \dfrac{3}{{2x + 6}} = \dfrac{3}{{2.6 + 6}} = \dfrac{3}{{18}} = \dfrac{1}{6}\\
b)A = \dfrac{{x - 3}}{{x + 3}} + \dfrac{{3x}}{{x - 3}} + \dfrac{{4{x^2} + 9}}{{9 - {x^2}}}\\
= \dfrac{{x - 3}}{{x + 3}} + \dfrac{{3x}}{{x - 3}} - \dfrac{{4{x^2} + 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{{\left( {x - 3} \right)}^2} + 3x\left( {x + 3} \right) - 4{x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 6x + 9 + 3{x^2} + 9x - 4{x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow C = \dfrac{A}{B}\\
= \dfrac{{3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\dfrac{3}{{2x + 6}}\\
= \dfrac{{3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2\left( {x + 3} \right)}}{3}\\
= \dfrac{{2x}}{{x - 3}}\\
c)Dkxd:x \ne 3;x \ne - 3\\
C > 2\\
\Rightarrow \dfrac{{2x}}{{x - 3}} > 2\\
\Rightarrow \dfrac{{2x}}{{x - 3}} - 2 > 0\\
\Rightarrow \dfrac{{2x - 2x + 6}}{{x - 3}} > 0\\
\Rightarrow \dfrac{6}{{x - 3}} > 0\\
\Rightarrow x - 3 > 0\\
\Rightarrow x > 3\\
Vay\,x > 3\\
B4)\\
a)Dkxd:x \ne 0;x \ne 3;x \ne - 3\\
P = \left( {\dfrac{{{x^2} - 3}}{{{x^2} - 9}} + \dfrac{1}{{x - 3}}} \right):\dfrac{x}{{x + 3}}\\
= \dfrac{{{x^2} - 3 + x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{x}\\
= \dfrac{{{x^2} + x}}{{x - 3}}.\dfrac{1}{x}\\
= \dfrac{{x + 1}}{{x - 3}}\\
b)\left| P \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
P = 3\\
P = - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\dfrac{{x + 1}}{{x - 3}} = 3\\
\dfrac{{x + 1}}{{x - 3}} = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 3x - 9\\
x + 1 = - 3x + 9
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
2x = 10\\
4x = 8
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 5\left( {tmdk} \right)\\
x = 2\left( {tmdk} \right)
\end{array} \right.
\end{array}$
