Đáp án:

a) \(\dfrac{{x - 3}}{{x + 1}}\)

b) Q=-3

c) \(m \ne \left\{ {0;3} \right\}\)

d) x=1

\(e)x >  - 1;x \ne \left\{ 3 \right\}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3; - 1;3} \right\}\\
Q = \dfrac{{x + 1 - x + 11}}{{x + 1}}:\dfrac{{{{\left( {x + 3} \right)}^2} + 36 - {{\left( {x - 3} \right)}^2}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{12}}{{x + 1}}:\dfrac{{{x^2} + 6x + 9 + 36 - {x^2} + 6x - 9}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{12}}{{x + 1}}.\dfrac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{12x + 36}}\\
 = \dfrac{{12}}{{x + 1}}.\dfrac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{12\left( {x + 3} \right)}}\\
 = \dfrac{{x - 3}}{{x + 1}}\\
b)2{x^2} + 6x = 0\\
 \to 2x\left( {x + 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
x =  - 3\left( l \right)
\end{array} \right.\\
Thay:x = 0\\
 \to Q = \dfrac{{0 - 3}}{{0 + 1}} =  - 3\\
c)Q = m\\
 \to \dfrac{{x - 3}}{{x + 1}} = m\\
 \to x - 3 = m\left( {x + 1} \right)\\
 \to x - 3 = mx + m\\
 \to \left( {m - 1} \right)x =  - m - 3\\
 \to x = \dfrac{{ - \left( {m + 3} \right)}}{{m - 1}}\\
Do:x \ne \left\{ { - 3; - 1;3} \right\}\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - m - 3}}{{m - 1}} \ne  - 3\\
\dfrac{{ - m - 3}}{{m - 1}} \ne  - 1\\
\dfrac{{ - m - 3}}{{m - 1}} \ne 3
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - m - 3 \ne  - 3m + 3\\
 - m - 3 \ne  - m + 1\\
 - m - 3 \ne 3m - 3
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
2m \ne 6\\
 - 3 \ne 1\\
4m \ne 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ne 3\\
m \ne 0
\end{array} \right.\\
 \to m \ne \left\{ {0;3} \right\}\\
d)Q =  - x\\
 \to \dfrac{{x - 3}}{{x + 1}} =  - x\\
 \to x - 3 =  - {x^2} - x\\
 \to {x^2} + 2x - 3 = 0\\
 \to \left( {x + 3} \right)\left( {x - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x =  - 3\left( l \right)\\
x = 1
\end{array} \right.\\
e)Q < 1\\
 \to \dfrac{{x - 3}}{{x + 1}} < 1\\
 \to \dfrac{{x - 3 - x - 1}}{{x + 1}} < 0\\
 \to \dfrac{{ - 4}}{{x + 1}} < 0\\
 \to x + 1 > 0\\
 \to x >  - 1;x \ne \left\{ 3 \right\}
\end{array}\)