Thực hiện các phép tính:
a) \({{{x^2} + x} \over {{x^2} - 16}}.{{4x - {x^2}} \over {3x + 3}}\) ;
b) \({{2{x^3} - 4{x^2}} \over {{x^2} + 8x + 16}}.{{3x + 12} \over {4x - {x^3}}}\) ;
c) \({{a + 3} \over {{a^2} - 2}}.{{8 - 12a + 6{a^2} - {a^3}} \over {9a + 27}}\)
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d) \({{x + 1} \over {{y^2} - 2y - 8}}.{{4 - y} \over {{x^2} + x}}\) .
\({1 \over x} + {{20} \over {17x}} = {{37} \over {17x}}\)
\(\eqalign{ & a)\,\,{{{x^2} + x} \over {{x^2} - 16}}.{{4x - {x^2}} \over {3x + 3}} = {{\left( {{x^2} + x} \right)\left( {4x - {x^2}} \right)} \over {\left( {{x^2} - 16} \right)\left( {3x + 3} \right)}} = {{x\left( {x + 1} \right)\left( { - x} \right)\left( {x - 4} \right)} \over {\left( {x - 4} \right)\left( {x + 4} \right)3\left( {x + 1} \right)}} = {{ - {x^2}} \over {3\left( {x + 4} \right)}} \cr & b)\,\,\,{{2{x^3} - 4{x^2}} \over {{x^2} + 8x + 16}}.{{3x + 12} \over {4x - {x^2}}} = {{\left( {2{x^3} - 4{x^2}} \right)\left( {3x + 12} \right)} \over {\left( {{x^2} + 8x + 16} \right)\left( {4x - {x^3}} \right)}} \cr & = {{ - 2{x^2}\left( {2 - x} \right)3\left( {x + 4} \right)} \over {{{\left( {x + 4} \right)}^2}x\left( {2 - x} \right)\left( {2 + x} \right)}} = {{ - 2x.3} \over {\left( {x + 4} \right)\left( {2 + x} \right)}} = {{ - 6x} \over {\left( {x + 4} \right)\left( {2 + x} \right)}} \cr & c)\,\,{{a + 3} \over {{a^2} - 4}}.{{8 - 12a + 6{a^2} - {a^3}} \over {9a + 27}} = {{\left( {a + 3} \right)\left( {8 - 12a + 6{a^2} - {a^3}} \right)} \over {\left( {{a^2} - 4} \right)\left( {9a + 27} \right)}} \cr & = {{\left( {a + 3} \right)\left( {{2^3} - {{3.2}^2}a + 3.2.{a^2} - {a^3}} \right)} \over {\left( {a - 2} \right)\left( {a + 2} \right)9\left( {a + 3} \right)}} = {{\left( {a + 3} \right){{\left( {2 - a} \right)}^3}} \over { - \left( {2 - a} \right)\left( {a + 2} \right)9\left( {a + 3} \right)}} \cr & = {{{{\left( {2 - a} \right)}^2}} \over { - 9\left( {a + 2} \right)}} = {{ - 2{{\left( {2 - a} \right)}^2}} \over {9\left( {a + 2} \right)}} \cr & d)\,\,{{x + 1} \over {{y^2} - 2y - 8}}.{{4 - y} \over {{x^2} + x}} = {{\left( {x + 1} \right)\left( {4 - y} \right)} \over {\left( {{y^2} - 2y - 8} \right)\left( {{x^2} + x} \right)}} \cr & = {{ - \left( {x + 1} \right)\left( {y - 4} \right)} \over {\left( {y - 4} \right)\left( {y + 2} \right)x\left( {x + 1} \right)}} = {{ - 1} \over {\left( {y + 2} \right)x}} \cr} \)