Advertisements (Quảng cáo)
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) \({5 \over {n + 6}} – {4 \over {n – 1}}\) ;
Advertisements (Quảng cáo)
d) \({{x – 5} \over {2x – 6}} – {{x – 7} \over {4x – 12}}\) .
\(\eqalign{ & a)\,\,{{4a + 2} \over {a + b}} + {1 \over { – a – b}} = {{4a + 2} \over {a + b}} + {{ – 1} \over {a + b}} = {{4a + 2 – 1} \over {a + b}} = {{4a + 1} \over {a + b}} \cr & b)\,\,{{2x} \over {5a{b^3}}} + {{4y} \over {3{a^2}{b^2}}} = {{2x.3a} \over {5a{b^3}.3a}} + {{4y.5b} \over {3{a^2}{b^2}.5b}} = {{6ax} \over {15{a^2}{b^3}}} + {{20by} \over {15{a^2}{b^3}}} = {{6ax + 20by} \over {15{a^2}{b^3}}} \cr & c)\,\,{5 \over {n + 6}} – {4 \over {n – 1}} = {{5\left( {n – 1} \right)} \over {\left( {n + 6} \right)\left( {n – 1} \right)}} – {{4\left( {n + 6} \right)} \over {\left( {n + 6} \right)\left( {n – 1} \right)}} = {{5n – 5 – 4n – 24} \over {\left( {n + 6} \right)\left( {n – 1} \right)}} = {{n – 29} \over {\left( {n + 6} \right)\left( {n – 1} \right)}} \cr & d)\,\,{{x – 5} \over {2x – 6}} – {{x – 7} \over {4x – 12}} = {{x – 5} \over {2\left( {x – 3} \right)}} + {{ – \left( {x – 7} \right)} \over {4\left( {x – 3} \right)}} \cr & = {{\left( {x – 5} \right).2} \over {2\left( {x – 3} \right).2}} + {{ – \left( {x – 7} \right)} \over {4\left( {x – 3} \right)}} = {{2x – 10 – x + 7} \over {4\left( {x – 3} \right)}} = {{x – 3} \over {4\left( {x – 3} \right)}} = {1 \over 4} \cr} \)