Rút gọn phân thức:
a) \({{3{x^3} + 6{x^2} + 3x} \over {{x^3} + 3{x^2} + 3x + 1}}\)
b) \({{x + 1} \over {{x^3} + 1}}\)
c) \({{{a^3} – {b^3}} \over {{a^2} – 2ab + {b^2}}}.\)
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a) \({{3{x^3} + 6{x^2} + 3x} \over {{x^3} + 3{x^2} + 3x + 1}} = {{3x\left( {{x^2} + 2x + 1} \right)} \over {{{\left( {x + 1} \right)}^3}}} = {{3x{{\left( {x + 1} \right)}^2}} \over {{{\left( {x + 1} \right)}^3}}}\)\(\; = {{3x} \over {x + 1}}.\)
Advertisements (Quảng cáo)
b) \({{x + 1} \over {{x^3} + 1}} = {{x + 1} \over {\left( {x + 1} \right)\left( {{x^2} – x + 1} \right)}} = {1 \over {{x^2} – x + 1}}.\)
c) \({{{a^3} – {b^3}} \over {{a^2} – 2ab + {b^2}}} = {{\left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)} \over {{{\left( {a – b} \right)}^2}}} = {{{a^2} + ab + {b^2}} \over {a – b}}.\)