Đáp án:

\(\begin{array}{l}
1,\\
a,\\
x\\
b,\\
\dfrac{x}{{3y}}\\
c,\\
\dfrac{{x + y}}{{3x}}\\
d,\\
\dfrac{{3.\left( {x + y} \right)}}{{x - y}}\\
2,\\
a,\\
\dfrac{{ - 1}}{{2x}}\\
b,\\
\dfrac{{x.\left( {{x^2} - 21} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}\\
c,\\
\dfrac{1}{{{x^2} + x + 1}}\\
d,\\
\dfrac{1}{{x - 1}}\\
3,\\
a,\\
DKXD:\,\,\left\{ \begin{array}{l}
x \ne  - 5\\
x \ne 0
\end{array} \right.\\
b,\\
B = \dfrac{{x - 1}}{2}\\
4,\\
A = \dfrac{x}{{x + 3}}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
\dfrac{{2{x^2} + 2xy}}{{2x + 2y}} = \dfrac{{x.\left( {2x + 2y} \right)}}{{2x + 2y}} = x\\
b,\\
\dfrac{{{x^2} - xy}}{{3xy - 3{y^2}}} = \dfrac{{x.\left( {x - y} \right)}}{{3y.\left( {x - y} \right)}} = \dfrac{x}{{3y}}\\
c,\\
\dfrac{{{x^2} + 2xy + {y^2}}}{{3{x^2} + 3xy}} = \dfrac{{{{\left( {x + y} \right)}^2}}}{{3x.\left( {x + y} \right)}} = \dfrac{{x + y}}{{3x}}\\
d,\\
\dfrac{{3{x^2} + 6xy + 3{y^2}}}{{{x^2} - {y^2}}} = \dfrac{{3.\left( {{x^2} + 2xy + {y^2}} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{3.{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{3.\left( {x + y} \right)}}{{x - y}}\\
2,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
x \ne \dfrac{1}{2}
\end{array} \right.\\
\dfrac{{1 - 3x}}{{2x}} + \dfrac{{3x - 2}}{{2x - 1}} + \dfrac{{3x - 2}}{{2x - 4{x^2}}}\\
 = \dfrac{{1 - 3x}}{{2x}} + \dfrac{{3x - 2}}{{2x - 1}} - \dfrac{{3x - 2}}{{4{x^2} - 2x}}\\
 = \dfrac{{1 - 3x}}{{2x}} + \dfrac{{3x - 2}}{{2x - 1}} - \dfrac{{3x - 2}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{\left( {1 - 3x} \right).\left( {2x - 1} \right) + 2x.\left( {3x - 2} \right) - \left( {3x - 2} \right)}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{\left( {1 - 3x} \right).\left( {2x - 1} \right) + \left( {3x - 2} \right).\left( {2x - 1} \right)}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{\left( {2x - 1} \right).\left[ {\left( {1 - 3x} \right) + \left( {3x - 2} \right)} \right]}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{\left( {2x - 1} \right).\left( {1 - 3x + 3x - 2} \right)}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{\left( {2x - 1} \right).\left( { - 1} \right)}}{{2x.\left( {2x - 1} \right)}}\\
 = \dfrac{{ - 1}}{{2x}}\\
b,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 3\\
x \ne  - 3
\end{array} \right.\\
\dfrac{1}{{{x^2} + 6x + 9}} + \dfrac{1}{{6x - {x^2} - 9}} + \dfrac{x}{{{x^2} - 9}}\\
 = \dfrac{1}{{{x^2} + 6x + 9}} - \dfrac{1}{{{x^2} - 6x + 9}} + \dfrac{x}{{{x^2} - 9}}\\
 = \dfrac{1}{{{x^2} + 2.x.3 + {3^2}}} - \dfrac{1}{{{x^2} - 2.x.3 + {3^2}}} + \dfrac{x}{{{x^2} - {3^2}}}\\
 = \dfrac{1}{{{{\left( {x + 3} \right)}^2}}} - \dfrac{1}{{{{\left( {x - 3} \right)}^2}}} + \dfrac{x}{{\left( {x - 3} \right).\left( {x + 3} \right)}}\\
 = \dfrac{{{{\left( {x - 3} \right)}^2} - {{\left( {x + 3} \right)}^2} + x.\left( {x - 3} \right).\left( {x + 3} \right)}}{{{{\left( {x - 3} \right)}^2}.{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{\left( {{x^2} - 6x + 9} \right) - \left( {{x^2} + 6x + 9} \right) + x.\left( {{x^2} - 9} \right)}}{{{{\left( {x - 3} \right)}^2}.{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{{x^2} - 6x + 9 - {x^2} - 6x - 9 + x.\left( {{x^2} - 9} \right)}}{{{{\left( {x - 3} \right)}^2}.{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{ - 12x + x.\left( {{x^2} - 9} \right)}}{{{{\left( {x - 3} \right)}^2}.{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{x.\left[ {\left( {{x^2} - 9} \right) - 12} \right]}}{{{{\left[ {\left( {x - 3} \right)\left( {x + 3} \right)} \right]}^2}}}\\
 = \dfrac{{x.\left( {{x^2} - 21} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}\\
c,\\
DKXD:\,\,\,x \ne 1\\
\dfrac{{{x^2} + 2}}{{{x^3} - 1}} + \dfrac{2}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}\\
 = \dfrac{{{x^2} + 2}}{{{x^3} - {1^3}}} + \dfrac{2}{{{x^2} + x + 1}} - \dfrac{1}{{x - 1}}\\
 = \dfrac{{{x^2} + 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{2}{{{x^2} + x + 1}} - \dfrac{1}{{x - 1}}\\
 = \dfrac{{{x^2} + 2 + 2.\left( {x - 1} \right) - \left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2 + 2x - 2 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{1}{{{x^2} + x + 1}}\\
d,\\
DKXD:\,\,\,x \ne 1\\
\dfrac{{6{x^2} + 8x + 7}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{6}{{1 - x}}\\
 = \dfrac{{6{x^2} + 8x + 7}}{{{x^3} - {1^3}}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{6}{{x - 1}}\\
 = \dfrac{{6{x^2} + 8x + 7}}{{{x^3} - {1^3}}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{6}{{x - 1}}\\
 = \dfrac{{6{x^2} + 8x + 7}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{6}{{x - 1}}\\
 = \dfrac{{6{x^2} + 8x + 7 + x\left( {x - 1} \right) - 6.\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{6{x^2} + 8x + 7 + {x^2} - x - 6{x^2} - 6x - 6}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{1}{{x - 1}}\\
3,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x + 10 \ne 0\\
x \ne 0\\
2x.\left( {x + 5} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \ne  - 10\\
x \ne 0\\
x + 5 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne  - 5\\
x \ne 0
\end{array} \right.\\
b,\\
B = \dfrac{{{x^2} + 2x}}{{2x + 10}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^2} + 2x}}{{2.\left( {x + 5} \right)}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{\left( {{x^2} + 2x} \right).x + \left( {x - 5} \right).2\left( {x + 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2.\left( {{x^2} - {5^2}} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2.\left( {{x^2} - 25} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x.\left( {x + 5} \right)}}\\
 = \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{x.\left( {{x^2} + 4x - 5} \right)}}{{2x.\left( {x + 5} \right)}}\\
 = \dfrac{{x.\left[ {\left( {{x^2} + 5x} \right) + \left( { - x - 5} \right)} \right]}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{x.\left[ {x.\left( {x + 5} \right) - \left( {x + 5} \right)} \right]}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{x.\left( {x + 5} \right).\left( {x - 1} \right)}}{{2x\left( {x + 5} \right)}}\\
 = \dfrac{{x - 1}}{2}\\
4,\\
DKXD:\,\,\,\,x \ne 1\\
A = \left( {\dfrac{{2{x^2} + 1}}{{{x^3} - 1}} - \dfrac{1}{{x - 1}}} \right):\left( {1 - \dfrac{{{x^2} - 2}}{{{x^2} + x + 1}}} \right)\\
 = \left( {\dfrac{{2{x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{1}{{x - 1}}} \right):\dfrac{{\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 2} \right)}}{{{x^2} + x + 1}}\\
 = \dfrac{{2{x^2} + 1 - \left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{{x^2} + x + 1 - {x^2} + 2}}{{{x^2} + x + 1}}\\
 = \dfrac{{2{x^2} + 1 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}:\dfrac{{x + 3}}{{{x^2} + x + 1}}\\
 = \dfrac{{{x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
 = \dfrac{{x.\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 3}}\\
 = \dfrac{x}{{x + 3}}
\end{array}\)