Đề bài

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) \({{8({x^2} - xy)} \over {12x{{(x - y)}^2}}}\) ;

d) \({{{x^2} + xy + x + y} \over {{x^2} - xy + x - y}}\) .

Lời giải chi tiết

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

dapandethi.vn