Giải thích các bước giải:
Bài 1:
\[\begin{array}{l}
a,\\
\frac{{5\left( {x - y} \right) - 3\left( {y - x} \right)}}{{10\left( {x - y} \right)}} = \frac{{\left( {x - y} \right)\left( {5 + 3} \right)}}{{10\left( {x - y} \right)}} = \frac{8}{{10}} = \frac{4}{5}\\
b,\\
\frac{{{x^2} + 4x + 3}}{{2x + 6}} = \frac{{\left( {{x^2} + 3x} \right) + \left( {x + 3} \right)}}{{2\left( {x + 3} \right)}} = \frac{{x\left( {x + 3} \right) + \left( {x + 3} \right)}}{{2\left( {x + 3} \right)}} = \frac{{\left( {x + 3} \right)\left( {x + 1} \right)}}{{2\left( {x + 3} \right)}} = \frac{{x + 1}}{2}\\
c,\\
\frac{{{y^2} - {x^2}}}{{{x^2} - 3xy + 2{y^2}}} = \frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{\left( {{x^2} - xy} \right) - \left( {2xy - 2{y^2}} \right)}}\\
= \frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{x\left( {x - y} \right) - 2y\left( {x - y} \right)}} = \frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{\left( {x - y} \right)\left( {x - 2y} \right)}} = \frac{{y + x}}{{2y - x}}\\
d,\\
\frac{{2{x^3} - 7{x^2} - 12x + 45}}{{3{x^3} - 19{x^2} + 33x - 9}} = \frac{{\left( {2{x^3} - 6{x^2}} \right) - \left( {{x^2} - 3x} \right) - \left( {15x - 45} \right)}}{{\left( {3{x^3} - 9{x^2}} \right) - \left( {10{x^2} - 30x} \right) + \left( {3x - 9} \right)}}\\
= \frac{{\left( {x - 3} \right)\left( {2{x^2} - x - 15} \right)}}{{\left( {x - 3} \right)\left( {3{x^2} - 10x + 3} \right)}} = \frac{{2{x^2} - x - 15}}{{3{x^2} - 10x + 3}} = \frac{{\left( {2x + 5} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {3x - 1} \right)}} = \frac{{2x + 5}}{{3x - 1}}
\end{array}\]
Bài 2:
\[\begin{array}{l}
A = \frac{{\left( {2{x^2} + 2x} \right){{\left( {x - 2} \right)}^2}}}{{\left( {{x^3} - 4x} \right)\left( {x + 1} \right)}} = \frac{{2x\left( {x + 1} \right){{\left( {x - 2} \right)}^2}}}{{x\left( {{x^2} - 4} \right)\left( {x + 1} \right)}}\\
= \frac{{2x\left( {x + 1} \right){{\left( {x - 2} \right)}^2}}}{{x\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right)}} = \frac{{2\left( {x - 2} \right)}}{{x + 2}}\\
x = \frac{1}{2} \Rightarrow A = - \frac{6}{5}\\
B = \frac{{{x^3} - {x^2}y + x{y^2}}}{{{x^3} + {y^3}}} = \frac{{x\left( {{x^2} - xy + {y^2}} \right)}}{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}} = \frac{x}{{x + y}}\\
x = - 5;y = 10 \Rightarrow B = - 1
\end{array}\]
