Đơn giản và rút gọn biểu thức:
a) \({{4a + 2} \over {a + b}} + {1 \over { - a - b}}\) ;
b) \({{2x} \over {5a{b^3}}} + {{4y} \over {3{a^2}{b^2}}}\) ;
c) \({5 \over {n + 6}} - {4 \over {n - 1}}\) ;
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d) \({{x - 5} \over {2x - 6}} - {{x - 7} \over {4x - 12}}\) .
\(\eqalign{ & a)\,\,{{4a + 2} \over {a + b}} + {1 \over { - a - b}} = {{4a + 2} \over {a + b}} + {{ - 1} \over {a + b}} = {{4a + 2 - 1} \over {a + b}} = {{4a + 1} \over {a + b}} \cr & b)\,\,{{2x} \over {5a{b^3}}} + {{4y} \over {3{a^2}{b^2}}} = {{2x.3a} \over {5a{b^3}.3a}} + {{4y.5b} \over {3{a^2}{b^2}.5b}} = {{6ax} \over {15{a^2}{b^3}}} + {{20by} \over {15{a^2}{b^3}}} = {{6ax + 20by} \over {15{a^2}{b^3}}} \cr & c)\,\,{5 \over {n + 6}} - {4 \over {n - 1}} = {{5\left( {n - 1} \right)} \over {\left( {n + 6} \right)\left( {n - 1} \right)}} - {{4\left( {n + 6} \right)} \over {\left( {n + 6} \right)\left( {n - 1} \right)}} = {{5n - 5 - 4n - 24} \over {\left( {n + 6} \right)\left( {n - 1} \right)}} = {{n - 29} \over {\left( {n + 6} \right)\left( {n - 1} \right)}} \cr & d)\,\,{{x - 5} \over {2x - 6}} - {{x - 7} \over {4x - 12}} = {{x - 5} \over {2\left( {x - 3} \right)}} + {{ - \left( {x - 7} \right)} \over {4\left( {x - 3} \right)}} \cr & = {{\left( {x - 5} \right).2} \over {2\left( {x - 3} \right).2}} + {{ - \left( {x - 7} \right)} \over {4\left( {x - 3} \right)}} = {{2x - 10 - x + 7} \over {4\left( {x - 3} \right)}} = {{x - 3} \over {4\left( {x - 3} \right)}} = {1 \over 4} \cr} \)