Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\rm{DK:}}\,\,\,\left\{ \begin{array}{l}
x \ne 3\\
x \ne - 3
\end{array} \right.\\
B = \left( {\frac{{3 - x}}{{x + 3}} + \frac{{{x^2} + 6x + 9}}{{{x^2} - 9}}} \right):\frac{{3{x^2}}}{{x + 3}}\\
= \left( {\frac{{3 - x}}{{x + 3}} + \frac{{{{\left( {x + 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right):\frac{{3{x^2}}}{{x + 3}}\\
= \left( {\frac{{3 - x}}{{x + 3}} + \frac{{x + 3}}{{x - 3}}} \right):\frac{{3{x^2}}}{{x + 3}}\\
= \left( {\frac{{\left( {3 - x} \right).\left( {x - 3} \right) + {{\left( {x + 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right):\frac{{3{x^2}}}{{x + 3}}\\
= \frac{{ - {x^2} + 6x - 9 + {x^2} + 6x + 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{3{x^2}}}\\
= \frac{{12x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{3{x^2}}}\\
= \frac{4}{{x\left( {x - 3} \right)}}\\
b,\\
{x^2} - 4x + 3 = 0 \Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3\left( L \right)
\end{array} \right.\\
x = 1 \Rightarrow B = \frac{4}{{1.\left( {1 - 3} \right)}} = - 2
\end{array}\)
\(\begin{array}{l}
c,\\
B > 0 \Leftrightarrow \frac{4}{{x\left( {x - 3} \right)}} > 0\\
\Leftrightarrow x\left( {x - 3} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x < 0
\end{array} \right.
\end{array}\)
