Thực hiện phép tính: a) 1/x^2 - x + 1:x + 1/x - 1 b) x + y/2x - y:1/x - y c) x^3y + xy^3/x^4y:( x^2 + y^2) d) x^3 + 8/x^2 - 2x + 1:x^2 + 3x + 2/1 - x^2

Lời giải

a) \(\frac{1}{{{x^2} - x + 1}}:\frac{{x + 1}}{{x - 1}} = \frac{1}{{{x^2} - x + 1}}.\frac{{x - 1}}{{x + 1}}\)

\( = \frac{{1.\left( {x - 1} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {x + 1} \right)}} = \frac{{x - 1}}{{{x^3} + 1}}\).

b) \(\frac{{x + y}}{{2x - y}}:\frac{1}{{x - y}} = \frac{{x + y}}{{2x - y}}.\frac{{x - y}}{1}\)

\( = \frac{{\left( {x + y} \right)\left( {x - y} \right)}}{{\left( {2x - y} \right).1}} = \frac{{{x^2} - {y^2}}}{{2x - y}}\).

c) \(\frac{{{x^3}y + x{y^3}}}{{{x^4}y}}:\left( {{x^2} + {y^2}} \right) = \frac{{{x^3}y + x{y^3}}}{{{x^4}y}}.\frac{1}{{{x^2} + {y^2}}}\)

\( = \frac{{\left( {{x^3}y + x{y^3}} \right).1}}{{{x^4}y.\left( {{x^2} + {y^2}} \right)}} = \frac{{xy.\left( {{x^2} + {y^2}} \right)}}{{{x^4}y.\left( {{x^2} + {y^2}} \right)}} = \frac{1}{{{x^3}}}\).

d) \(\frac{{{x^3} + 8}}{{{x^2} - 2x + 1}}:\frac{{{x^2} + 3x + 2}}{{1 - {x^2}}}\)

\( = \frac{{{x^3} + 8}}{{{x^2} - 2x + 1}}.\frac{{1 - {x^2}}}{{{x^2} + 3x + 2}}\)

\( = \frac{{\left( {{x^3} + 8} \right)\left( {1 - {x^2}} \right)}}{{\left( {{x^2} - 2x + 1} \right)\left( {{x^2} + 3x + 2} \right)}}\)

\( = \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {1 - x} \right)\left( {1 + x} \right)}}{{{{\left( {x - 1} \right)}^2}.\left[ {\left( {{x^2} - 1} \right) + \left( {3x + 3} \right)} \right]}}\)

\( = \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {1 - x} \right)\left( {1 + x} \right)}}{{{{\left( {1 - x} \right)}^2}.\left[ {\left( {x - 1} \right)\left( {x + 1} \right) + 3.\left( {x + 1} \right)} \right]}}\)

\[ = \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {1 + x} \right)}}{{\left( {1 - x} \right).\left( {x + 1} \right)\left( {x - 1 + 3} \right)}}\]

\[ = \frac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {1 - x} \right).\left( {x + 2} \right)}}\]

\( = \frac{{{x^2} - 2x + 4}}{{1 - x}}\).