Câu 9 trang 26 Sách bài tập (SBT) Toán 8 tập 1
Rút gọn các phân thức sau:
a. ({{14x{y^5}left( {2x – 3y} right)} over {21{x^2}y{{left( {2x – 3y} right)}^2}}})
b. ({{8xy{{left( {3x – 1} right)}^3}} over {12{x^3}left( {1 – 3x} right)}})
c. ({{20{x^2} – 45} over {{{left( {2x + 3} right)}^2}}})
d.({{5{x^2} – 10xy} over {2{{left( {2y – x} right)}^3}}})
e. ({{80{x^3} – 125x} over {3left( {x – 3} right) – left( {x – 3} right)left( {8 – 4x} right)}})
f. ({{9 – {{left( {x + 5} right)}^2}} over {{x^2} + 4x + 4}})
g. ({{32x – 8{x^2} + 2{x^3}} over {{x^3} + 64}})
h. ({{5{x^3} + 5x} over {{x^4} – 1}})
i. ({{{x^2} + 5x + 6} over {{x^2} + 4x + 4}})
Giải:
a. ({{14x{y^5}left( {2x – 3y} right)} over {21{x^2}y{{left( {2x – 3y} right)}^2}}}) (= {{2{y^4}} over {3xleft( {2x – 3y} right)}})
b. ({{8xy{{left( {3x – 1} right)}^3}} over {12{x^3}left( {1 – 3x} right)}}) ( = {{ – 8xy{{left( {3x – 1} right)}^3}} over {12{x^2}left( {3x – 1} right)}} = {{ – 2y{{left( {3x – 1} right)}^2}} over {3x}})
c. ({{20{x^2} – 45} over {{{left( {2x + 3} right)}^2}}}) ( = {{5left( {4{x^2} – 9} right)} over {{{left( {2x + 3} right)}^2}}} = {{5left( {2x + 3} right)left( {2x – 3} right)} over {{{left( {2x + 3} right)}^2}}} = {{5left( {2x – 3} right)} over {2x + 3}})
d. ({{5{x^2} – 10xy} over {2{{left( {2y – x} right)}^3}}}) ( = {{ – 5xleft( {2y – x} right)} over {2{{left( {2y – x} right)}^3}}} = {{ – 5x} over {2{{left( {2y – x} right)}^2}}})
e. ({{80{x^3} – 125x} over {3left( {x – 3} right) – left( {x – 3} right)left( {8 – 4x} right)}}) ( = {{5xleft( {16{x^2} – 25} right)} over {left( {x – 3} right)left( {3 – 8 + 4x} right)}} = {{5xleft( {16{x^2} – 25} right)} over {left( {x – 3} right)left( {4x – 5} right)}} = {{5xleft( {4x + 5} right)} over {x – 3}})
f. ({{9 – {{left( {x + 5} right)}^2}} over {{x^2} + 4x + 4}}) ( = {{left( {3 + x + 5} right)left( {3 – x – 5} right)} over {{{left( {x + 2} right)}^2}}} = {{ – left( {8 + x} right)left( {x + 2} right)} over {{{left( {x + 2} right)}^2}}} = {{ – left( {8 + x} right)} over {x + 2}})
g. ({{32x – 8{x^2} + 2{x^3}} over {{x^3} + 64}}) ( = {{2xleft( {16 – 4x + {x^2}} right)} over {left( {x + 4} right)left( {{x^2} – 4x + 16} right)}} = {{2x} over {x + 4}})
h. ({{5{x^3} + 5x} over {{x^4} – 1}})( = {{5xleft( {{x^2} + 1} right)} over {left( {{x^2} – 1} right)left( {{x^2} + 1} right)}} = {{5x} over {{x^2} – 1}})
i. ({{{x^2} + 5x + 6} over {{x^2} + 4x + 4}}) ( = {{{x^2} + 2x + 3x + 6} over {{{left( {x + 2} right)}^2}}} = {{xleft( {x + 2} right) + 3left( {x + 2} right)} over {{{left( {x + 2} right)}^2}}} = {{left( {x + 2} right)left( {x + 3} right)} over {{{left( {x + 2} right)}^2}}} = {{x + 3} over {x + 2}})