a) \(\displaystyle {{3x} \over {x - 5}} = {{ - \left( {3x} \right)} \over { - \left( {x - 5} \right)}} = {{ - 3x} \over {5 - x}}\)
\(\displaystyle {{7x + 2} \over {5 - x}}\)
b) \(\displaystyle {{4x} \over {x + 1}} = {{4x\left( {x - 1} \right)} \over {\left( {x + 1} \right)\left( {x - 1} \right)}}\)\(\,\displaystyle = {{4{x^2} - 4x} \over {{x^2} - 1}}\)
\(\displaystyle {{3x} \over {x - 1}}= {{3x\left( {x + 1} \right)} \over {\left( {x - 1} \right)\left( {x + 1} \right)}} \)\(\,\displaystyle = {{3{x^2} + 3x} \over {{x^2} - 1}}\)
c) \(\displaystyle {2 \over {{x^2} + 8x + 16}}=\frac{2}{{{x^2} + 2.x.4 + {4^2}}} \)
\(\displaystyle = \frac{2}{{{{\left( {x + 4} \right)}^2}}} = {4 \over {2{{\left( {x + 4} \right)}^2}}}\)
\(\displaystyle {{x - 4} \over {2x + 8}} = {{x - 4} \over {2\left( {x + 4} \right)}} \)\(\,\displaystyle = {{\left( {x - 4} \right)\left( {x + 4} \right)} \over {2\left( {x + 4} \right)\left( {x + 4} \right)}} = {{{x^2} - 16} \over {2{{\left( {x + 4} \right)}^2}}}\)
d. \(\displaystyle {{2x} \over {\left( {x + 1} \right)\left( {x - 3} \right)}} \)\(\,\displaystyle = {{2x\left( {x - 2} \right)} \over {\left( {x + 1} \right)\left( {x - 3} \right)\left( {x - 2} \right)}}\)\(\,\displaystyle = {{2{x^2} - 4x} \over {\left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\)
\(\displaystyle {{x + 3} \over {\left( {x + 1} \right)\left( {x - 2} \right)}}\)\(\,\displaystyle = {{\left( {x + 3} \right)\left( {x - 3} \right)} \over {\left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\)\(\,\displaystyle = {{{x^2} - 9} \over {\left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}\)
