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Thực hiện các phép tính sau bằng hai cách : dùng tính chất phân phối của phép nhân đối với phép cộng và không dùng tính chất này :
a. \({{{x^3} – 1} \over {x + 2}}.\left( {{1 \over {x – 1}} – {{x + 1} \over {{x^2} + x + 1}}} \right)\)
b. \({{{x^3} + 2{x^2} – x – 2} \over {2x + 10}}\left( {{1 \over {x – 1}} – {2 \over {x + 1}} + {1 \over {x + 2}}} \right)\)
Cách 1 :
a. \({{{x^3} – 1} \over {x + 2}}.\left( {{1 \over {x – 1}} – {{x + 1} \over {{x^2} + x + 1}}} \right)\)
\(\eqalign{ & = {{{x^3} – 1} \over {x + 2}}.{1 \over {x – 1}} – {{{x^3} – 1} \over {x + 2}}.{{x + 1} \over {{x^2} + x + 1}} \cr & = {{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)} \over {\left( {x + 2} \right)\left( {x – 1} \right)}} – {{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)} \over {\left( {x + 2} \right)\left( {{x^2} + x + 1} \right)}} \cr & = {{{x^2} + x + 1} \over {x + 2}} – {{{x^2} – 1} \over {x + 2}} = {{{x^2} + x + 1 – {x^2} + 1} \over {x + 2}} = {{x + 2} \over {x + 2}} = 1 \cr} \)
Cách 2 : \({{{x^3} – 1} \over {x + 2}}.\left( {{1 \over {x – 1}} – {{x + 1} \over {{x^2} + x + 1}}} \right)\)
\(\eqalign{ & = {{{x^3} – 1} \over {x + 2}}.\left[ {{{{x^2} + x + 1} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} – {{\left( {x + 1} \right)\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}} \right] \cr & = {{{x^3} – 1} \over {x + 2}}.{{{x^2} + x + 1 – {x^2} + 1} \over {{x^3} – 1}} = {{{x^3} – 1} \over {x + 2}}.{{x + 2} \over {{x^3} – 1}} = 1 \cr} \)
b.
Cách 1 : \({{{x^3} + 2{x^2} – x – 2} \over {2x + 10}}\left( {{1 \over {x – 1}} – {2 \over {x + 1}} + {1 \over {x + 2}}} \right)\)
\(\eqalign{ & = {{{x^2}\left( {x + 2} \right) – \left( {x + 2} \right)} \over {2x + 10}}.\left( {{1 \over {x – 1}} – {2 \over {x + 1}} + {1 \over {x + 2}}} \right) \cr & = {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{1 \over {x – 1}} – {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{2 \over {x + 1}} + {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{1 \over {x + 2}} \cr & = {{\left( {x + 2} \right)\left( {x + 1} \right)} \over {2\left( {x + 5} \right)}} – {{2\left( {x + 2} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}} + {{\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}} \cr & = {{{x^2} + 2x + x + 2 – 2{x^2} + 2x – 4x + 4 + {x^2} – 1} \over {2\left( {x + 5} \right)}} = {{x + 5} \over {2\left( {x + 5} \right)}} = {1 \over 2} \cr} \)
Cách 2 : \({{{x^3} + 2{x^2} – x – 2} \over {2x + 10}}\left( {{1 \over {x – 1}} – {2 \over {x + 1}} + {1 \over {x + 2}}} \right)\)
\(\eqalign{ & = {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{{\left( {x – 1} \right)\left( {x + 2} \right) – 2\left( {x – 1} \right)\left( {x + 2} \right) + \left( {x + 1} \right)\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}} \cr & = {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{{{x^2} + 2x + x + 2 – 2{x^2} – 4x + 2x + 4 + {x^2} – 1} \over {\left( {x – 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)}} \cr & = {{\left( {x + 2} \right)\left( {x + 1} \right)\left( {x – 1} \right)} \over {2\left( {x + 5} \right)}}.{{x + 5} \over {\left( {x + 1} \right)\left( {x – 1} \right)\left( {x + 2} \right)}} = {1 \over 2} \cr} \)