a) \(\displaystyle A = 12{x^2} + 9x = 3x\left( {4x + 3} \right)\)
\(\displaystyle \Rightarrow {{4x + 3} \over {{x^2} - 5}} = {{\left( {4x + 3} \right).3x} \over {\left( {{x^2} - 5} \right).3x}} \)\(\displaystyle \,= {{12{x^2} + 9x} \over {3{x^3} - 15x}}\)
b) Ta có:
\(\begin{array}{l}8{x^2} - 8x + 2\\ = 2\left( {4{x^2} - 4x + 1} \right)\\ = 2\left[ {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right]\\ = 2{\left( {2x - 1} \right)^2} = 2{\left( {1 - 2x} \right)^2}\end{array}\)
\(\, \Rightarrow (8{x^2} - 8x + 2):(1 - 2x )\)\(\,=2\left( {1 - 2x} \right) = 2 - 4x\)
\(\displaystyle {{8{x^2} - 8x + 2} \over {\left( {4x - 2} \right)\left( {15 - x} \right)}} \)
\(\displaystyle= {{\left( {8{x^2} - 8x + 2} \right):\left( {2 - 4x} \right)} \over {\left( {4x - 2} \right)\left( {15 - x} \right):\left( {2 - 4x} \right)}}\)
\(\displaystyle = {{1 - 2x} \over {x - 15}}\)
