Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\dfrac{{2x - 2y}}{{3y - 3x}} = \dfrac{{2.\left( {x - y} \right)}}{{3\left( {y - x} \right)}} = \dfrac{{2\left( {x - y} \right)}}{{ - 3.\left( {x - y} \right)}} = \dfrac{2}{{ - 3}} = - \dfrac{2}{3}\\
b,\\
\dfrac{{{x^2} - x + 1}}{x} = \dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right).x}} = \dfrac{{{x^3} + {1^3}}}{{{x^2} + x}} = \dfrac{{{x^3} + 1}}{{{x^2} + x}}\\
2,\\
a,\\
\dfrac{{x + 3}}{{{x^2} + 4x + 3}} = \dfrac{{x + 3}}{{\left( {{x^2} + 3x} \right) + \left( {x + 3} \right)}} = \dfrac{{x + 3}}{{x\left( {x + 3} \right) + \left( {x + 3} \right)}} = \dfrac{{x + 3}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{1}{{x + 1}}\\
\dfrac{{x + 5}}{{{x^2} + 6x + 5}} = \dfrac{{x + 5}}{{\left( {{x^2} + 5x} \right) + \left( {x + 5} \right)}} = \dfrac{{x + 5}}{{x\left( {x + 5} \right) + \left( {x + 5} \right)}} = \dfrac{{x + 5}}{{\left( {x + 5} \right)\left( {x + 1} \right)}} = \dfrac{1}{{x + 1}}\\
\Rightarrow \dfrac{{x + 3}}{{{x^2} + 4x + 3}} = \dfrac{{x + 5}}{{{x^2} + 6x + 5}}\\
b,\\
{\left( {x - y} \right)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3} \ne {x^3} - {y^3}\\
\Rightarrow \dfrac{1}{{{{\left( {x - y} \right)}^3}}} \ne \dfrac{1}{{{x^3} - {y^3}}}\\
3,\\
a,\\
\dfrac{{{x^2}{y^3}}}{{2{x^2}{y^2}}} = \dfrac{1}{2}.\dfrac{{{x^2}}}{{{x^2}}}.\dfrac{{{y^3}}}{{{y^2}}} = \dfrac{1}{2}.1.y = \dfrac{y}{2} \ne \dfrac{y}{x}\\
b,\\
\dfrac{{x\left( {x - 2} \right)}}{{2\left( {x - 2} \right)}} = \dfrac{x}{2}\\
c,\\
\dfrac{{x - 2}}{{{x^2} - 4}} = \dfrac{{x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x + 2}} \ne \dfrac{1}{{x - 2}}\\
d,\\
\dfrac{{1 - x}}{{\left( {x - 1} \right)\left( {3 - x} \right)}} = \dfrac{{ - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left[ { - \left( {x - 3} \right)} \right]}} = \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
e,\\
\dfrac{{ - 2\left( {x - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}} = \dfrac{{ - 2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} \ne \dfrac{{2.\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\)
