Advertisements (Quảng cáo)
Dùng quy tắc biến đổi dấu rồi thực hiện các phép tính:
a)\({{4x + 13} \over {5x\left( {x – 7} \right)}} – {{x – 48} \over {5x\left( {7 – x} \right)}};\)
b)\({1 \over {x – 5{x^2}}} – {{25 – 15} \over {25{x^2} – 1}}\)
Hướng dẫn làm bài:
a) \({{4x + 13} \over {5x\left( {x – 7} \right)}} – {{x – 48} \over {5x\left( {7 – x} \right)}} = {{4x + 13} \over {5x\left( {x – 7} \right)}} + {{x – 48} \over { – 5x\left( {7 – x} \right)}}\)
\( = {{4x + 13} \over {5x\left( {x – 7} \right)}} + {{x – 48} \over {5x\left( {x – 7} \right)}} = {{4x + 13 + x – 48} \over {5x\left( {x – 7} \right)}}\)
\( = {{5x – 35} \over {5x\left( {x – 7} \right)}} = {{5\left( {x – 7} \right)} \over {5x\left( {x – 7} \right)}} = {1 \over x}\)
Advertisements (Quảng cáo)
b) \({1 \over {x – 5{x^2}}} – {{25 – 15} \over {25{x^2} – 1}} = {1 \over {x\left( {1 – 5x} \right)}} + {{25x – 15} \over { – \left( {25{x^2} – 1} \right)}}\)
\( = {1 \over {x\left( {1 – 5x} \right)}} + {{25x – 15} \over {1 – 25{x^2}}} = {1 \over {x\left( {1 – 5x} \right)}} + {{25x – 15} \over {\left( {1 – 5x} \right)\left( {1 + 5x} \right)}}\)
\( = {{1 + 5x + x\left( {25x – 15} \right)} \over {x\left( {1 – 5x} \right)\left( {1 + 5x} \right)}} = {{1 + 5x + 25{x^2} – 15} \over {x\left( {1 – 5x} \right)\left( {1 + 5x} \right)}}\)
\( = {{1 – 10x + 25{x^2}} \over {x\left( {1 – 5x} \right)\left( {1 + 5x} \right)}} = {{{{\left( {1 – 5x} \right)}^2}} \over {x\left( {1 – 5x} \right)\left( {1 + 5x} \right)}} = {{1 – 5x} \over {x\left( {1 + 5x} \right)}}\)