Rút gọn các phân thức sau:
a) \({{8x(1 – x)} \over {12{x^2}{{(x – 1)}^3}}}\) ;
b) \({{10x{{(2 – x)}^3}} \over {25{x^3}{{(x – 2)}^2}}}\).
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\(\eqalign{ & a)\,\,{{8x\left( {1 – x} \right)} \over {12{x^2}{{\left( {x – 1} \right)}^3}}} = {{ – 8x\left( {x – 1} \right)} \over {12{x^2}{{\left( {x – 1} \right)}^3}}} \cr & \,\,\,\,\, = {{\left[ {4x\left( {x – 1} \right)} \right].\left( { – 2} \right)} \over {\left[ {4x\left( {x – 1} \right)} \right].\left[ {3x{{\left( {x – 1} \right)}^2}} \right]}} = {{ – 2} \over {3x{{\left( {x – 1} \right)}^2}}} \cr & b)\,\,{{10x{{\left( {2 – x} \right)}^3}} \over {25{x^3}{{\left( {x – 2} \right)}^2}}} = {{10x{{\left( {2 – x} \right)}^3}} \over {25{x^3}{{\left( {2 – x} \right)}^2}}} \cr & \,\,\,\,\, = {{\left[ {5x{{\left( {2 – x} \right)}^2}} \right].\left[ {2\left( {2 – x} \right)} \right]} \over {\left[ {5x{{\left( {2 – x} \right)}^2}} \right].\left( {5{x^2}} \right)}} = {{2\left( {2 – x} \right)} \over {5{x^2}}} \cr} \)