Câu 29 trang 32 Sách bài tập (SBT) Toán 8 tập 1
Làm tính nhân phân thức :
a. ({{30{x^3}} over {11{y^2}}}.{{121{y^5}} over {25x}})
b. ({{24{y^5}} over {7{x^2}}}.left( { – {{21x} over {12{y^3}}}} right))
c. (left( { – {{18{y^3}} over {25{x^4}}}} right).left( { – {{15{x^2}} over {9{y^3}}}} right))
d. ({{4x + 8} over {{{left( {x – 10} right)}^3}}}.{{2x – 20} over {{{left( {x + 2} right)}^2}}})
e. ({{2{x^2} – 20x + 50} over {3x + 3}}.{{{x^2} – 1} over {4{{left( {x – 5} right)}^3}}})
Giải:
a. ({{30{x^3}} over {11{y^2}}}.{{121{y^5}} over {25x}})( = {{30{x^3}.121{y^5}} over {11{y^2}.25x}} = {{6{x^2}.11{y^3}} over {1.5}} = {{66{x^2}{y^3}} over 5})
b. ({{24{y^5}} over {7{x^2}}}.left( { – {{21x} over {12{y^3}}}} right)) ( = {{24{y^5}.left( { – 21x} right)} over {7{x^2}.12{y^3}}} = {{2{y^2}.left( { – 3} right)} over x} = – {{6{y^2}} over x})
c. (left( { – {{18{y^3}} over {25{x^4}}}} right).left( { – {{15{x^2}} over {9{y^3}}}} right)) ( = {{left( { – 18{y^3}} right).left( { – 15{x^2}} right)} over {25{x^4}.9{y^3}}} = {{ – 2.left( { – 3} right)} over {5{x^2}.1}} = {6 over {5{x^2}}})
d. ({{4x + 8} over {{{left( {x – 10} right)}^3}}}.{{2x – 20} over {{{left( {x + 2} right)}^2}}})( = {{4left( {x + 2} right).2left( {x – 10} right)} over {{{left( {x – 10} right)}^3}{{left( {x + 2} right)}^2}}} = {8 over {{{left( {x – 10} right)}^2}left( {x + 2} right)}})
e. ({{2{x^2} – 20x + 50} over {3x + 3}}.{{{x^2} – 1} over {4{{left( {x – 5} right)}^3}}})( = {{2left( {{x^2} – 10x + 25} right)left( {x + 1} right)left( {x – 1} right)} over {3left( {x + 1} right).4{{left( {x – 5} right)}^3}}})
( = {{{{left( {x – 5} right)}^2}left( {x – 1} right)} over {6{{left( {x – 5} right)}^3}}} = {{x – 1} over {6left( {x – 5} right)}})