Đáp án:

\(\begin{array}{l}
1,\\
\dfrac{{ - 1}}{{x\left( {x + 2} \right)}}\\
2,\\
\dfrac{{x - 6}}{{4\left( {x - 1} \right)}}\\
3,\\
\dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\\
4,\\
{x^2} - {y^2}\\
5,\\
\dfrac{{5x + 5}}{{4x + 12}}\\
6,\\
\dfrac{{6x - 6}}{{7x - 56}}\\
7,\\
\dfrac{{4x + 4}}{{5x + 30}}\\
8,\\
\dfrac{{3x + 3}}{{5x - 35}}\\
9,\\
\dfrac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\\
10,\\
\dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\\
11,\\
\dfrac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 3} \right)}^2}}}
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
\dfrac{{x + 1}}{{{x^2} - 2x - 8}}.\dfrac{{4 - x}}{{{x^2} + x}}\\
 = \dfrac{{x + 1}}{{\left( {{x^2} - 4x} \right) + \left( {2x - 8} \right)}}.\dfrac{{4 - x}}{{x\left( {x + 1} \right)}}\\
 = \dfrac{{x + 1}}{{x\left( {x - 4} \right) + 2\left( {x - 4} \right)}}.\dfrac{{4 - x}}{{x\left( {x + 1} \right)}}\\
 = \dfrac{{x + 1}}{{\left( {x - 4} \right)\left( {x + 2} \right)}}.\dfrac{{4 - x}}{{x\left( {x + 1} \right)}}\\
 = \dfrac{1}{{x + 2}}.\dfrac{{ - 1}}{x}\\
 = \dfrac{{ - 1}}{{x\left( {x + 2} \right)}}\\
2,\\
\dfrac{{x + 2}}{{4x + 24}}.\dfrac{{{x^2} - 36}}{{{x^2} + x - 2}}\\
 = \dfrac{{x + 2}}{{4\left( {x + 6} \right)}}.\dfrac{{{x^2} - {6^2}}}{{\left( {{x^2} - x} \right) + \left( {2x - 2} \right)}}\\
 = \dfrac{{x + 2}}{{4\left( {x + 6} \right)}}.\dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{x\left( {x - 1} \right) + 2\left( {x - 1} \right)}}\\
 = \dfrac{{x + 2}}{{4\left( {x + 6} \right)}}.\dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\\
 = \dfrac{1}{4}.\dfrac{{x - 6}}{{x - 1}}\\
 = \dfrac{{x - 6}}{{4\left( {x - 1} \right)}}\\
3,\\
\dfrac{{{x^2} - 5x + 6}}{{{x^2} + 7x + 12}}:\dfrac{{{x^2} - 4x + 4}}{{{x^2} + 3x}}\\
 = \dfrac{{\left( {{x^2} - 3x} \right) + \left( { - 2x + 6} \right)}}{{\left( {{x^2} + 3x} \right) + \left( {4x + 12} \right)}}:\dfrac{{{x^2} - 2.x.2 + {2^2}}}{{x\left( {x + 3} \right)}}\\
 = \dfrac{{x\left( {x - 3} \right) - 2.\left( {x - 3} \right)}}{{x\left( {x + 3} \right) + 4\left( {x + 3} \right)}}:\dfrac{{{{\left( {x - 2} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
 = \dfrac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}.\dfrac{{x\left( {x + 3} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
 = \dfrac{{x - 3}}{{x + 4}}.\dfrac{x}{{x - 2}}\\
 = \dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x + 4} \right)}}\\
4,\\
\dfrac{{{x^2} - 2xy + {y^2}}}{{{x^2} - xy + {y^2}}}:\dfrac{{x - y}}{{{x^3} + {y^3}}}\\
 = \dfrac{{{x^2} - 2.x.y + {y^2}}}{{{x^2} - xy + {y^2}}}:\dfrac{{x - y}}{{\left( {x + y} \right)\left( {{x^2} - x.y + {y^2}} \right)}}\\
 = \dfrac{{{{\left( {x - y} \right)}^2}}}{{{x^2} - xy + {y^2}}}.\dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{\left( {x - y} \right)}}\\
 = \dfrac{{x - y}}{1}.\dfrac{{x + y}}{1}\\
 = \left( {x - y} \right)\left( {x + y} \right)\\
 = {x^2} - {y^2}\\
5,\\
\dfrac{{5x - 15}}{{4x + 4}}:\dfrac{{{x^2} - 9}}{{{x^2} + 2x + 1}}\\
 = \dfrac{{5.\left( {x - 3} \right)}}{{4\left( {x + 1} \right)}}:\dfrac{{{x^2} - {3^2}}}{{{x^2} + 2.x.1 + {1^2}}}\\
 = \dfrac{{5.\left( {x - 3} \right)}}{{4\left( {x + 1} \right)}}:\dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
 = \dfrac{{5.\left( {x - 3} \right)}}{{4\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{5}{4}.\dfrac{{x + 1}}{{x + 3}}\\
 = \dfrac{{5x + 5}}{{4x + 12}}\\
6,\\
\dfrac{{6x + 48}}{{7x - 7}}:\dfrac{{{x^2} - 64}}{{{x^2} - 2x + 1}}\\
 = \dfrac{{6.\left( {x + 8} \right)}}{{7\left( {x - 1} \right)}}:\dfrac{{{x^2} - {8^2}}}{{{x^2} - 2.x.1 + {1^2}}}\\
 = \dfrac{{6.\left( {x + 8} \right)}}{{7\left( {x - 1} \right)}}:\dfrac{{\left( {x - 8} \right)\left( {x + 8} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{6.\left( {x + 8} \right)}}{{7.\left( {x - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 8} \right)\left( {x + 8} \right)}}\\
 = \dfrac{6}{7}.\dfrac{{x - 1}}{{x - 8}}\\
 = \dfrac{{6x - 6}}{{7x - 56}}\\
7,\\
\dfrac{{4x - 24}}{{5x + 5}}:\dfrac{{{x^2} - 36}}{{{x^2} + 2x + 1}}\\
 = \dfrac{{4.\left( {x - 6} \right)}}{{5\left( {x + 1} \right)}}:\dfrac{{{x^2} - {6^2}}}{{{x^2} + 2.x.1 + {1^2}}}\\
 = \dfrac{{4.\left( {x - 6} \right)}}{{5.\left( {x + 1} \right)}}:\dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
 = \dfrac{{4.\left( {x - 6} \right)}}{{5\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 6} \right)\left( {x + 6} \right)}}\\
 = \dfrac{4}{5}.\dfrac{{x + 1}}{{x + 6}}\\
 = \dfrac{{4x + 4}}{{5x + 30}}\\
8,\\
\dfrac{{3x + 21}}{{5x + 5}}:\dfrac{{{x^2} - 49}}{{{x^2} + 2x + 1}}\\
 = \dfrac{{3.\left( {x + 7} \right)}}{{5.\left( {x + 1} \right)}}:\dfrac{{{x^2} - {7^2}}}{{{x^2} + 2x + 1}}\\
 = \dfrac{{3.\left( {x + 7} \right)}}{{5\left( {x + 1} \right)}}:\dfrac{{\left( {x - 7} \right)\left( {x + 7} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
 = \dfrac{{3\left( {x + 7} \right)}}{{5\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 7} \right)\left( {x + 7} \right)}}\\
 = \dfrac{3}{5}.\dfrac{{x + 1}}{{x - 7}}\\
 = \dfrac{{3x + 3}}{{5x - 35}}\\
9,\\
\dfrac{{x + 1}}{{x + 2}}:\dfrac{{x + 2}}{{x + 3}}:\dfrac{{x + 3}}{{x + 1}}\\
 = \dfrac{{x + 1}}{{x + 2}}.\dfrac{{x + 3}}{{x + 2}}.\dfrac{{x + 1}}{{x + 3}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\\
10,\\
\dfrac{{x + 1}}{{x + 2}}:\left( {\dfrac{{x + 2}}{{x + 3}}:\dfrac{{x + 3}}{{x + 1}}} \right)\\
 = \dfrac{{x + 1}}{{x + 2}}:\left( {\dfrac{{x + 2}}{{x + 3}}.\dfrac{{x + 1}}{{x + 3}}} \right)\\
 = \dfrac{{x + 1}}{{x + 2}}:\dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{x + 1}}{{x + 2}}.\dfrac{{{{\left( {x + 3} \right)}^2}}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\\
11,\\
\dfrac{{x + 1}}{{x + 2}}.\dfrac{{x + 2}}{{x + 3}}:\dfrac{{x + 3}}{{x + 1}}\\
 = \dfrac{{x + 1}}{{x + 2}}.\dfrac{{x + 2}}{{x + 3}}.\dfrac{{x + 1}}{{x + 3}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 3} \right)}^2}}}
\end{array}\)