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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\(\eqalign{ & a)\,\,{{{x^2} + {y^2}} \over {{x^2} – 2xy + {y^2}}}.{{x – y} \over {{x^2}}} – {{2{y^2}} \over {{x^2} – 2xy + {y^2}}}.{{x – y} \over {{x^2}}} \cr & = \left( {{{{x^2} + {y^2}} \over {{x^2} – 2xy + {y^2}}} – {{2{y^2}} \over {{x^2} – 2xy + {y^2}}}} \right){{x – y} \over {{x^2}}} \cr & = {{{x^2} + {y^2} – 2{y^2}} \over {{{\left( {x – y} \right)}^2}}}.{{x – y} \over {{x^2}}} = {{{x^2} – {y^2}} \over {x – y}}.{1 \over {{x^2}}} \cr & = {{\left( {x – y} \right)\left( {x + y} \right)} \over {x – y}}.{1 \over {{x^2}}} = {{x + y} \over {{x^2}}} \cr & b)\,\,\left( {{{{x^3}} \over {x – 1}} + {x^2} – x + 1} \right).{{1 – x} \over x} \cr & = {{{x^3}} \over {x – 1}}.{{1 – x} \over x} + {{{x^2}\left( {1 – x} \right)} \over x} – {{x\left( {1 – x} \right)} \over x} + {{1\left( {1 – x} \right)} \over x} \cr & = {{ – {x^2}\left( {x – 1} \right)} \over {x – 1}} + x\left( {1 – x} \right) – \left( {1 – x} \right) + {{1 – x} \over x} \cr & = – {x^2} + x – {x^2} – 1 + x + {{1 – x} \over x} \cr & = – 2{x^2} + 2x – 1 + {{1 – x} \over x} \cr} \)