Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{20{x^2} - 45}}{{{{\left( {2x + 3} \right)}^2}}} = \dfrac{{5.\left( {4{x^2} - 9} \right)}}{{{{\left( {2x + 3} \right)}^2}}} = \dfrac{{5.\left[ {{{\left( {2x} \right)}^2} - {3^2}} \right]}}{{{{\left( {2x + 3} \right)}^2}}} = \dfrac{{5.\left( {2x - 3} \right)\left( {2x + 3} \right)}}{{{{\left( {2x + 3} \right)}^2}}} = \dfrac{{5.\left( {2x - 3} \right)}}{{2x + 3}}\\
b,\\
\dfrac{{{x^3} - 3{x^2} - x + 3}}{{{x^2} - 3x}} = \dfrac{{\left( {{x^3} - 3{x^2}} \right) + \left( { - x + 3} \right)}}{{x\left( {x - 3} \right)}} = \dfrac{{{x^2}\left( {x - 3} \right) - \left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} = \dfrac{{\left( {x - 3} \right)\left( {{x^2} - 1} \right)}}{{x\left( {x - 3} \right)}} = \dfrac{{{x^2} - 1}}{x}\\
c,\\
\dfrac{{{x^2} + 7x + 12}}{{{x^2} + 5x + 6}} = \dfrac{{\left( {{x^2} + 3x} \right) + \left( {4x + 12} \right)}}{{\left( {{x^2} + 2x} \right) + \left( {3x + 6} \right)}} = \dfrac{{x.\left( {x + 3} \right) + 4.\left( {x + 3} \right)}}{{x\left( {x + 2} \right) + 3.\left( {x + 2} \right)}} = \dfrac{{\left( {x + 3} \right)\left( {x + 4} \right)}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{x + 4}}{{x + 2}}\\
d,\\
\dfrac{{{x^2} - {y^2}}}{{{x^2} - {y^2} + xz - yz}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right) + z.\left( {x - y} \right)}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y + z} \right)}} = \dfrac{{x + y}}{{x + y + z}}\\
e,\\
\dfrac{{2x + 3 - {x^2}}}{{{x^2} - 1}} = \dfrac{{\left( { - {x^2} - x} \right) + \left( {3x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{ - x\left( {x + 1} \right) + 3.\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{\left( {x + 1} \right)\left( { - x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{ - x + 3}}{{x - 1}}
\end{array}\)
