Đáp án:

a) \(\dfrac{{6{x^2} + 3x + 18}}{{{x^2} + 3x - 18}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne  \pm 3\\
P = \left( {\dfrac{x}{{x - 3}} - \dfrac{2}{{x + 3}} - \dfrac{{{x^2}}}{{9 - {x^2}}}} \right):\dfrac{{x + 6}}{{3\left( {x + 3} \right)}}\\
 = \left[ {\dfrac{{x\left( {x + 3} \right) - 2\left( {x - 3} \right) + {x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right].\dfrac{{3\left( {x + 3} \right)}}{{x + 6}}\\
 = \dfrac{{{x^2} + 3x - 2x + 6 + {x^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{3\left( {x + 3} \right)}}{{x + 6}}\\
 = \dfrac{{2{x^2} + x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{3\left( {x + 3} \right)}}{{x + 6}}\\
 = \dfrac{{3\left( {2{x^2} + x + 6} \right)}}{{\left( {x - 3} \right)\left( {x + 6} \right)}} = \dfrac{{6{x^2} + 3x + 18}}{{{x^2} + 3x - 18}}\\
b)2x - \left| {4 - x} \right| = 5\\
 \to \left| {4 - x} \right| = 2x - 5\\
 \to \left[ \begin{array}{l}
4 - x = 2x - 5\\
4 - x =  - 2x + 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3x = 7\\
x = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = 1
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
P = \dfrac{{6.{{\left( {\dfrac{7}{3}} \right)}^2} + 3.\left( {\dfrac{7}{3}} \right) + 18}}{{{{\left( {\dfrac{7}{3}} \right)}^2} + 3.\left( {\dfrac{7}{3}} \right) - 18}} = \dfrac{{ - 519}}{{50}}\\
P = \dfrac{{{{6.1}^2} + 3.1 + 18}}{{{1^2} + 3.1 - 18}} =  - \dfrac{{27}}{{14}}
\end{array} \right.\\
c)P = \dfrac{{6{x^2} + 3x + 18}}{{{x^2} + 3x - 18}}
\end{array}\)

( bạn xem lại đề có nhầm số và dấu ở đâu không nhé )