a) Vì \(a ≠ \displaystyle \pm {3 \over 2}\) nên \(\displaystyle4{a^2} - 9 \ne 0 \) \(\displaystyle\Rightarrow x = {{4a + 4} \over {4{a^2} - 9}}\)
Vì \(a ≠ − 1\) nên \(\displaystyle3{a^3} + a \ne 0 \) \(\displaystyle \Rightarrow y = {{6{a^2} + 9a} \over {3{a^3} + 3}}\)
Do đó:
\(\displaystyle xy = {{4a + 4} \over {4{a^2} - 9}}.{{6{a^2} + 9a} \over {3{a^3} + 3}} \) \(\displaystyle = {{4\left( {a + 1} \right).3a\left( {2a + 3} \right)} \over {\left( {2a + 3} \right)\left( {2a - 3} \right).3\left( {{a^3} + 1} \right)}}\)
\(\displaystyle = {{4a\left( {a + 1} \right)} \over {\left( {2a - 3} \right)\left( {a + 1} \right)\left( {{a^2} - a + 1} \right)}} \) \(\displaystyle= {{4a} \over {\left( {2a - 3} \right)\left( {{a^2} - a + 1} \right)}}\)
b) Vì \(a ≠ b\) nên \(\displaystyle 2{a^3} - 2{b^3} \ne 0 \) \(\displaystyle \Rightarrow x = {{3a + 3b} \over {2{a^3} - 2{b^3}}}\)
Vì \(a ≠ − b\) nên \(\displaystyle 6a + 6b \ne 0 \) \(\displaystyle \Rightarrow y = {{{{\left( {a - b} \right)}^2}} \over {6a + 6b}}\)
Do đó :
\(\displaystyle xy = {{3a + 3b} \over {2{a^3} - 2{b^3}}}.{{{{\left( {a - b} \right)}^2}} \over {6a + 6b}} \) \(\displaystyle = {{3\left( {a + b} \right){{\left( {a - b} \right)}^2}} \over {2\left( {{a^3} - {b^3}} \right).6\left( {a + b} \right)}}\)
\(\displaystyle = {{{{\left( {a - b} \right)}^2}} \over {4\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)}} \) \(\displaystyle= {{a - b} \over {4\left( {{a^2} + ab + {b^2}} \right)}}\)
