Đáp án:

\(\begin{array}{l}
a,\,\,\,x \ne 0\\
b,\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
x \ne 1
\end{array} \right.\\
c,\,\,\,x \ne 2\\
d,\,\,x \ne 2\\
e,\,\,x \ne 1\\
2,\\
1.\,\,\,a\,\,\,\,\,\,\,\\
2.\,\,\,\dfrac{a}{{b - a}}\,\,\,\,\\
3.\,\,\,x - y\,\,\,\,\,\\
4.\,\,\,\dfrac{{3x}}{{2x - 1}}\,\,\,\,\\
5.\,\,\,\,\dfrac{1}{{y - x}}\,\,\,\,\,\\
6.\,\,\,\dfrac{{x - y}}{{x + y}}\,\,\,\,\,\,\,\\
3,\\
1,\,\,\,\dfrac{{2{x^2} - 3x + 9}}{{\left( {x + 3} \right){{\left( {x - 3} \right)}^2}}}\\
2,\,\,\,\,\dfrac{{ - {x^2} + 6x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
3,\,\,\,\, - 1\\
4,\,\,\,\,\dfrac{x}{{15}}\\
5,\,\,\,\,\dfrac{1}{2}\\
6,\,\,\,\,\dfrac{x}{{6{y^2}}}\\
7,\,\,\,\,\dfrac{{ - 10}}{{3{y^2}}}\\
8,\,\,\,\, - \dfrac{{2x + 1}}{{2x - 1}}\\
9,\,\,\,\,2
\end{array}\)

Giải thích các bước giải:

 Ta có:

Bài 1:

Các phân số đã cho xác định khi và chỉ khi:

\(\begin{array}{l}
a,\,\,\,x \ne 0\\
b,\,\,\,x\left( {x - 1} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 1
\end{array} \right.\\
c,\,\,\,5x - 10 \ne 0 \Leftrightarrow 5.\left( {x - 2} \right) \ne 0 \Leftrightarrow x - 2 \ne 0 \Leftrightarrow x \ne 2\\
e,\,\,x - 1 \ne 0 \Leftrightarrow x \ne 1\\
2,\\
1.\,\,\,\dfrac{{{a^2} - ab}}{{a - b}} = \dfrac{{a.\left( {a - b} \right)}}{{a - b}} = a\,\,\,\,\,\,\,\left( {a \ne b} \right)\\
2.\,\,\,\dfrac{{{a^2}b}}{{a{b^2} - {a^2}b}} = \dfrac{{a.ab}}{{ab.b - a.ab}} = \dfrac{{a.ab}}{{ab.\left( {b - a} \right)}} = \dfrac{a}{{b - a}}\,\,\,\,\left( {a \ne b \ne 0} \right)\\
3.\,\,\,\dfrac{{{x^2} - 2xy + {y^2}}}{{x - y}} = \dfrac{{{{\left( {x - y} \right)}^2}}}{{x - y}} = x - y\,\,\,\,\,\left( {x \ne y} \right)\\
4.\,\,\,\dfrac{{3x + 6{x^2}}}{{4{x^2} - 1}} = \dfrac{{3x.\left( {1 + 2x} \right)}}{{{{\left( {2x} \right)}^2} - {1^2}}} = \dfrac{{3x\left( {1 + 2x} \right)}}{{\left( {2x - 1} \right)\left( {2x + 1} \right)}} = \dfrac{{3x}}{{2x - 1}}\,\,\,\,\left( {x \ne  \pm \dfrac{1}{2}} \right)\\
5.\,\,\,\,\dfrac{{y - x}}{{{x^2} - 2xy + {y^2}}} = \dfrac{{y - x}}{{{y^2} - 2.y.x + {x^2}}} = \dfrac{{y - x}}{{{{\left( {y - x} \right)}^2}}} = \dfrac{1}{{y - x}}\,\,\,\,\,\left( {x \ne y} \right)\\
6.\,\,\,\dfrac{{{x^2} - xy - x + y}}{{{x^2} + xy - x - y}} = \dfrac{{\left( {{x^2} - xy} \right) + \left( { - x + y} \right)}}{{\left( {{x^2} + xy} \right) + \left( { - x - y} \right)}}\\
 = \dfrac{{x.\left( {x - y} \right) - \left( {x - y} \right)}}{{x\left( {x + y} \right) - \left( {x + y} \right)}} = \dfrac{{\left( {x - y} \right)\left( {x - 1} \right)}}{{\left( {x + y} \right)\left( {x - 1} \right)}} = \dfrac{{x - y}}{{x + y}}\,\,\,\,\,\,\,\left( \begin{array}{l}
x \ne 1\\
x \ne  - y
\end{array} \right)\\
3,\\
1.\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ne 3\\
x \ne  - 3
\end{array} \right.\\
\dfrac{1}{{x + 3}} + \dfrac{x}{{{x^2} - 6x + 9}} = \dfrac{{{x^2} - 6x + 9 + x.\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {{x^2} - 6x + 9} \right)}}\\
 = \dfrac{{{x^2} - 6x + 9 + {x^2} + 3x}}{{\left( {x + 3} \right)\left( {{x^2} - 2.x.3 + {3^2}} \right)}} = \dfrac{{2{x^2} - 3x + 9}}{{\left( {x + 3} \right){{\left( {x - 3} \right)}^2}}}\\
2,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ne 3\\
x \ne  - 3
\end{array} \right.\\
\dfrac{{2x}}{{{x^2} - 9}} - \dfrac{{x - 1}}{{x + 3}} = \dfrac{{2x}}{{{x^2} - {3^2}}} - \dfrac{{x - 1}}{{x + 3}}\\
 = \dfrac{{2x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{x - 1}}{{x + 3}} = \dfrac{{2x - \left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{2x - \left( {{x^2} - 4x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{2x - {x^2} + 4x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{ - {x^2} + 6x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
3,\\
DKXD:\left\{ \begin{array}{l}
x \ne 2\\
x \ne  - \dfrac{1}{2}
\end{array} \right.\\
\dfrac{{2x + 1}}{{x - 2}}.\dfrac{{2 - x}}{{2x + 1}} = \dfrac{{2x + 1}}{{x - 2}}.\dfrac{{ - \left( {x - 2} \right)}}{{2x + 1}} =  - 1\\
4,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0\\
x \ne \dfrac{{ - 2}}{7}
\end{array} \right.\\
\dfrac{{7x + 2}}{{5x{y^3}}}.\dfrac{{{x^2}{y^3}}}{{21x + 6}} = \dfrac{{7x + 2}}{{5.x{y^3}}}.\dfrac{{x.x{y^3}}}{{3.\left( {7x + 2} \right)}} = \dfrac{1}{5}.\dfrac{x}{3} = \dfrac{x}{{15}}\\
5,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 3
\end{array} \right.\\
\dfrac{{{x^2} + 6x + 9}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{2{x^2} - 4x + 2}}{{4{x^2} + 24x + 36}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{2\left( {{x^2} - 2x + 1} \right)}}{{4.\left( {{x^2} + 6x + 9} \right)}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{2\left( {{x^2} - 2.x.1 + {1^2}} \right)}}{{4.\left( {{x^2} + 6x + 9} \right)}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{2{{\left( {x - 1} \right)}^2}}}{{4.\left( {{x^2} + 6x + 9} \right)}}\\
 = \dfrac{2}{4} = \dfrac{1}{2}\\
6,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0\\
x \ne  - \dfrac{2}{7}
\end{array} \right.\\
\dfrac{{7x + 2}}{{3x{y^3}}}:\dfrac{{14x + 4}}{{{x^2}y}} = \dfrac{{7x + 2}}{{3x{y^3}}}.\dfrac{{{x^2}y}}{{14x + 4}}\\
 = \dfrac{{7x + 2}}{{xy.3{y^2}}}.\dfrac{{xy.x}}{{2.\left( {7x + 2} \right)}} = \dfrac{1}{{3{y^2}}}.\dfrac{x}{2} = \dfrac{x}{{6{y^2}}}\\
7,\\
DKXD:\,\,\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0\\
x \ne \dfrac{1}{3}
\end{array} \right.\\
\dfrac{{8xy}}{{3x - 1}}:\dfrac{{12x{y^3}}}{{5 - 15x}} = \dfrac{{8xy}}{{3x - 1}}.\dfrac{{5 - 15x}}{{12x{y^3}}}\\
 = \dfrac{{2.4xy}}{{3x - 1}}.\dfrac{{ - 5.\left( {3x - 1} \right)}}{{3{y^2}.4xy}} = \dfrac{2}{1}.\dfrac{{ - 5}}{{3{y^2}}} = \dfrac{{ - 10}}{{3{y^2}}}\\
8,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 2\\
x \ne \dfrac{1}{2}
\end{array} \right.\\
\dfrac{{2x + 1}}{{x - 2}}:\left( { - \dfrac{{2x - 1}}{{x - 2}}} \right) = \dfrac{{2x + 1}}{{x - 2}}.\left( { - \dfrac{{x - 2}}{{2x - 1}}} \right)\\
 = \left( {2x + 1} \right).\left( { - \dfrac{1}{{2x - 1}}} \right) =  - \dfrac{{2x + 1}}{{2x - 1}}\\
9,\\
DKXD:\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 2
\end{array} \right.\\
\dfrac{{{x^2} + 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}:\dfrac{{2{x^2} + 4x + 2}}{{4{x^2} - 8x + 4}}\\
 = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{4{x^2} - 8x + 4}}{{2{x^2} + 4x + 2}}\\
 = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{4.\left( {{x^2} - 2x + 1} \right)}}{{2.\left( {{x^2} + 2x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{4.\left( {{x^2} - 2.x.1 + {1^2}} \right)}}{{2.\left( {{x^2} + 2x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{4{{\left( {x - 1} \right)}^2}}}{{2.\left( {{x^2} + 2x + 1} \right)}}\\
 = \dfrac{4}{2} = 2
\end{array}\)