Câu 3.1 trang 27 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn phân thức :

a. \(\dfrac{{{x^4} - {y^4}}}{{{y^3} - {x^3}}}\)

b. \(\dfrac{{\left( {2x - 4} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {3{x^2} - 27} \right)}}\)

c. \(\dfrac{{2{x^3} + {x^2} - 2x - 1}}{{{x^3} + 2{x^2} - x - 2}}\)

Giải:

a. 

\(\eqalign{
& \frac{{{x^4} - {y^4}}}{{{y^3} - {x^3}}} \cr
& = \frac{{\left( {{x^2} + {y^2}} \right)\left( {{x^2} - {y^2}} \right)}}{{\left( {y - x} \right)\left( {{y^2} + xy + {x^2}} \right)}} \cr
& = \frac{{\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)\left( {x - y} \right)}}{{\left( {y - x} \right)\left( {{y^2} + xy + {x^2}} \right)}} \cr
& = - \frac{{\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)\left( {x - y} \right)}}{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}} \cr
& = - \frac{{\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}{{{x^2} + xy + {y^2}}} \cr} \)

b. 

\(\eqalign{
& \frac{{\left( {2x - 4} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {3{x^2} - 27} \right)}} \cr
& = \frac{{2\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 2} \right).3.\left( {{x^2} - 9} \right)}} \cr
& = \frac{{2\left( {x - 3} \right)}}{{3\left( {x + 3} \right)\left( {x - 3} \right)}} \cr
& = \frac{2}{{3\left( {x + 3} \right)}} \cr} \)

c. 

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

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