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Rút gọn phân thức
a) \({{25(x – 2)} \over {20x(2 – x)}}\) ;
b) \({{x{{(4 – x)}^2}} \over {x – 4}}\) ;
c) \({{{{(x – y)}^2}} \over {x{{(y – x)}^3}}}\) ;
Advertisements (Quảng cáo)
d) \({{x(x – 2)} \over {{{(2 – x)}^3}}}\) .
\(\eqalign{ & a)\,\,{{25\left( {x – 2} \right)} \over {20x\left( {2 – x} \right)}} = {{25\left( {x – 2} \right)} \over { – 20x\left( {x – 2} \right)}} \cr & \,\,\,\,\, = {{5\left[ {5\left( {x – 2} \right)} \right]} \over { – 4x\left[ {5\left( {x – 2} \right)} \right]}} = {5 \over { – 4x}} \cr & b)\,\,{{x{{\left( {4 – x} \right)}^2}} \over {x – 4}} = {{x{{\left( {4 – x} \right)}^2}} \over {\left( {4 – x} \right)\left( { – 1} \right)}} \cr & \,\,\,\,\, = {{x\left( {4 – x} \right)} \over { – 1}} = {x^2} – 4x \cr & c)\,\,{{{{\left( {x – y} \right)}^2}} \over {x{{\left( {y – x} \right)}^3}}} = {{{{\left( {y – x} \right)}^2}} \over {x{{\left( {y – x} \right)}^3}}} \cr & \,\,\,\,\, = {{{{\left( {y – x} \right)}^2}} \over {{{\left( {y – x} \right)}^2}\left[ {x\left( {y – x} \right)} \right]}} = {1 \over {x\left( {y – x} \right)}} \cr & d)\,\,{{x\left( {x – 2} \right)} \over {{{\left( {2 – x} \right)}^3}}} = {{ – x\left( {2 – x} \right)} \over {{{\left( {2 – x} \right)}^3}}} \cr & \,\,\,\,\, = {{ – x\left( {2 – x} \right)} \over {\left( {2 – x} \right){{\left( {2 – x} \right)}^2}}} = {{ – x} \over {{{\left( {2 – x} \right)}^2}}} \cr} \)