Đề bài

Thực hiện phép tính: a) \(\frac{1}{{{x^2} - x + 1}}:\frac{{x + 1}}{{x - 1}}\) b) \(\frac{{x + y}}{{2x - y}}:\frac{1}{{x - y}}\) c) \(\frac{{{x^3}y + x{y^3}}}{{{x^4}y}}:\left( {{x^2} + {y^2}} \right)\) d) \(\frac{{{x^3} + 8}}{{{x^2} - 2x + 1}}:\frac{{{x^2} + 3x + 2}}{{1 - {x^2}}}\)

Lời giải chi tiết

a) \(\frac{1}{{{x^2} - x + 1}}:\frac{{x + 1}}{{x - 1}} = \frac{1}{{{x^2} - x + 1}}.\frac{{x - 1}}{{x + 1}} = \frac{{x - 1}}{{{x^3} + 1}}\)b) \(\frac{{x + y}}{{2x - y}}:\frac{1}{{x - y}} = \frac{{x + y}}{{2x - y}}.\frac{{x - y}}{1} = \frac{{{x^2} - {y^2}}}{{2x - y}}\)c) \(\frac{{{x^3}y + x{y^3}}}{{{x^4}y}}:\left( {{x^2} + {y^2}} \right) = \frac{{xy\left( {{x^2} + {y^2}} \right)}}{{{x^4}y}}.\frac{1}{{{x^2} + {y^2}}} = \frac{1}{{{x^3}}}\)d) \(\frac{{{x^3} + 8}}{{{x^2} - 2x + 1}}:\frac{{{x^2} + 3x + 2}}{{1 - {x^2}}} = \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + {y^2}} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\frac{{ - \left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 2} \right)\left( {x + 1} \right)}} =  - \frac{{{x^2} - 2x + 4}}{{x - 1}}\)