\[Q = 1 + \frac{{x + 3}}{{{x^2} + 5x + 6}}:\left( {\frac{{8{x^2}}}{{4{x^3} - 8{x^2}}} - \frac{{3x}}{{3{x^2} - 12}} - \frac{1}{{x + 2}}} \right)\]

a) Rút gọn Q;

b) Tìm x để Q = 0;

c) Tìm x để Q > 0.

a) \[Q = 1 + \frac{{x + 3}}{{{x^2} + 5x + 6}}:\left( {\frac{{8{x^2}}}{{4{x^3} - 8{x^2}}} - \frac{{3x}}{{3{x^2} - 12}} - \frac{1}{{x + 2}}} \right)\]

\[ = 1 + \frac{{x + 3}}{{{x^2} + 3x + 2x + 6}}:\left[ {\frac{{8{x^2}}}{{4{x^2}\left( {x - 2} \right)}} - \frac{{3x}}{{3\left( {{x^2} - 4} \right)}} - \frac{{x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]\]

\[ = 1 + \frac{{x + 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)}}:\left[ {\frac{2}{{x - 2}} - \frac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \frac{{x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}} \right]\]

\[ = 1 + \frac{1}{{x + 2}}:\left[ {\frac{{2\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{{x + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]\]

\[ = 1 + \frac{1}{{x + 2}}:\left[ {\frac{{2x + 4 - 2x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]\]

\[ = 1 + \frac{1}{{x + 2}}:\frac{6}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\]

\[ = 1 + \frac{1}{{x + 2}}.\frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{6}\]

\[ = 1 + \frac{{x - 2}}{6} = \frac{{x + 4}}{6}\].

b) Khi Q = 0 thì \[\frac{{x + 4}}{6} = 0\]

Suy ra x + 4 = 0

Do đó x = - 4

Vậy x = - 4 khi Q = 0.

c) Khi Q > 0 thì \[\frac{{x + 4}}{6} > 0\]

Suy ra x + 4 > 0

Do đó x > - 4

Vậy x > - 4 thì Q > 0.