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Rút gọn phân thức
a) \({{6{x^3}{y^2}} \over {9{x^2}y}}\) ;
b) \({{6x{y^3}} \over {4{x^2}y}}\) ;
c) \({{8({x^2} – xy)} \over {12x{{(x – y)}^2}}}\) ;
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d) \({{{x^2} + xy + x + y} \over {{x^2} – xy + x – y}}\) .
\(\eqalign{ & a)\,\,{{6{x^3}{y^2}} \over {9{x^2}y}} = {{\left( {3{x^2}y} \right)\left( {2xy} \right)} \over {\left( {3{x^2}y} \right).3}} = {{2xy} \over 3} \cr & b)\,\,{{6x{y^3}} \over {4{x^2}y}} = {{\left( {2xy} \right)\left( {3{y^2}} \right)} \over {\left( {2xy} \right)\left( {2x} \right)}} = {{3{y^2}} \over {2x}} \cr & c)\,\,{{8\left( {{x^2} – xy} \right)} \over {12x{{\left( {x – y} \right)}^2}}} = {{8x\left( {x – y} \right)} \over {12x{{\left( {x – y} \right)}^2}}} \cr & \,\,\,\,\,\, = {{\left[ {4x\left( {x – y} \right)} \right].2} \over {\left[ {4x\left( {x – y} \right)} \right].\left[ {3\left( {x – y} \right)} \right]}} = {2 \over {3\left( {x – y} \right)}} \cr & d)\,\,{{{x^2} + xy + x + y} \over {{x^2} – xy + x – y}} = {{\left( {{x^2} + xy} \right) + \left( {x + y} \right)} \over {\left( {{x^2} – xy} \right) + \left( {x – y} \right)}} \cr & \,\,\,\,\, = {{x\left( {x + y} \right) + \left( {x + y} \right)} \over {x\left( {x – y} \right) + \left( {x – y} \right)}} = {{\left( {x + y} \right)\left( {x + 1} \right)} \over {\left( {x – y} \right)\left( {x + 1} \right)}} = {{x + y} \over {x – y}} \cr} \)