Áp dụng tính chất phân phối để thực hiện phép tính:
a) \({{{x^2} + {y^2}} \over {{x^2} - 2xy + {y^2}}}.{{x - y} \over {{x^2}}} - {{2{y^2}} \over {{x^2} - 2xy + {y^2}}}.{{x - y} \over {{x^2}}}\) ;
b) \(\left( {{{{x^3}} \over {x - 1}} + {x^2} - x + 1} \right).{{1 - x} \over x}\) .
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\(\eqalign{ & a)\,\,{{{x^2} + {y^2}} \over {{x^2} - 2xy + {y^2}}}.{{x - y} \over {{x^2}}} - {{2{y^2}} \over {{x^2} - 2xy + {y^2}}}.{{x - y} \over {{x^2}}} \cr & = \left( {{{{x^2} + {y^2}} \over {{x^2} - 2xy + {y^2}}} - {{2{y^2}} \over {{x^2} - 2xy + {y^2}}}} \right){{x - y} \over {{x^2}}} \cr & = {{{x^2} + {y^2} - 2{y^2}} \over {{{\left( {x - y} \right)}^2}}}.{{x - y} \over {{x^2}}} = {{{x^2} - {y^2}} \over {x - y}}.{1 \over {{x^2}}} \cr & = {{\left( {x - y} \right)\left( {x + y} \right)} \over {x - y}}.{1 \over {{x^2}}} = {{x + y} \over {{x^2}}} \cr & b)\,\,\left( {{{{x^3}} \over {x - 1}} + {x^2} - x + 1} \right).{{1 - x} \over x} \cr & = {{{x^3}} \over {x - 1}}.{{1 - x} \over x} + {{{x^2}\left( {1 - x} \right)} \over x} - {{x\left( {1 - x} \right)} \over x} + {{1\left( {1 - x} \right)} \over x} \cr & = {{ - {x^2}\left( {x - 1} \right)} \over {x - 1}} + x\left( {1 - x} \right) - \left( {1 - x} \right) + {{1 - x} \over x} \cr & = - {x^2} + x - {x^2} - 1 + x + {{1 - x} \over x} \cr & = - 2{x^2} + 2x - 1 + {{1 - x} \over x} \cr} \)