a) \(\dfrac{{14x{y^5}\left( {2x - 3y} \right)}}{{21{x^2}y{{\left( {2x - 3y} \right)}^2}}} = \dfrac{{2{y^4}}}{{3x\left( {2x - 3y} \right)}}\)
b) \(\dfrac{{8xy{{\left( {3x - 1} \right)}^3}}}{{12{x^3}\left( {1 - 3x} \right)}} = \dfrac{{8xy{{\left( {3x - 1} \right)}^3}}}{{ - 12{x^2}\left( {3x - 1} \right)}}\)
\(= \dfrac{{ - 8xy{{\left( {3x - 1} \right)}^3}}}{{12{x^2}\left( {3x - 1} \right)}} = \dfrac{{ - 2y{{\left( {3x - 1} \right)}^2}}}{{3x}}\)
c) \(\displaystyle {{20{x^2} - 45} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {4{x^2} - 9} \right)} \over {{{\left( {2x + 3} \right)}^2}}} \)
\(\displaystyle = {{5\left( {2x + 3} \right)\left( {2x - 3} \right)} \over {{{\left( {2x + 3} \right)}^2}}} = {{5\left( {2x - 3} \right)} \over {2x + 3}} \)
d) \(\dfrac{{5{x^2} - 10xy}}{{2{{\left( {2y - x} \right)}^3}}} = \dfrac{{ - 5x\left( {2y - x} \right)}}{{2{{\left( {2y - x} \right)}^3}}}\)
\( = \dfrac{{ - 5x}}{{2{{\left( {2y - x} \right)}^2}}}\)
e) \(\displaystyle {{80{x^3} - 125x} \over {3\left( {x - 3} \right) - \left( {x - 3} \right)\left( {8 - 4x} \right)}} \)
\( = \dfrac{{5x\left( {16{x^2} - 25} \right)}}{{\left( {x - 3} \right)\left[ {3 - \left( {8 - 4x} \right)} \right]}} \)
\(= \dfrac{{5x\left[ {{{\left( {4x} \right)}^2} - {5^2}} \right]}}{{\left( {x - 3} \right)\left( {3 - 8 + 4x} \right)}}\)
\(\displaystyle = {{5x\left( {4x - 5} \right)\left( {4x + 5} \right)} \over {\left( {x - 3} \right)\left( {4x - 5} \right)}} \)
\(\displaystyle = {{5x\left( {4x + 5} \right)} \over {x - 3}} \)
f) \( \displaystyle{{9 - {{\left( {x + 5} \right)}^2}} \over {{x^2} + 4x + 4}} = \dfrac{{{3^2} - {{\left( {x + 5} \right)}^2}}}{{{x^2} + 2.x.2 + {2^2}}}\)
\(\displaystyle = {{\left( {3 + x + 5} \right)\left( {3 - x - 5} \right)} \over {{{\left( {x + 2} \right)}^2}}} \)
\(\displaystyle= {{\left( {8 + x} \right)\left( { - 2 - x} \right)} \over {{{\left( {x + 2} \right)}^2}}} \)
\(\displaystyle = {{ - \left( {8 + x} \right)\left( {x + 2} \right)} \over {{{\left( {x + 2} \right)}^2}}} = {{ - 8- x } \over {x + 2}} \)
g) \( \displaystyle {{32x - 8{x^2} + 2{x^3}} \over {{x^3} + 64}} \)
\( = \dfrac{{32x - 8{x^2} + 2{x^3}}}{{{x^3} + {4^3}}}\)
\(\displaystyle = {{2x\left( {16 - 4x + {x^2}} \right)} \over {\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right)}} \)
\(\displaystyle = {{2x} \over {x + 4}} \)
h) \( \displaystyle {{5{x^3} + 5x} \over {{x^4} - 1}} = {{5x\left( {{x^2} + 1} \right)} \over {\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} \)
\(\displaystyle = {{5x} \over {{x^2} - 1}} \)
i) \( \dfrac{{{x^2} + 5x + 6}}{{{x^2} + 4x + 4}} \)
\(= \dfrac{{{x^2} + 2x + 3x + 6}}{{{x^2} + 2.x.2 + {2^2}}} \)
\(= \dfrac{{x\left( {x + 2} \right) + 3\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}}} \)
\(= \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{{{\left( {x + 2} \right)}^2}}} = \dfrac{{x + 3}}{{x + 2}} \)
