Câu 29 trang 32 Sách bài tập (SBT) Toán 8 tập 1

Làm tính nhân phân thức :

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}}}} ight))

c. (left( { - {{18{y^3}} over {25{x^4}}}} ight).left( { - {{15{x^2}} over {9{y^3}}}} ight))

d. ({{4x + 8} over {{{left( {x - 10} ight)}^3}}}.{{2x - 20} over {{{left( {x + 2} ight)}^2}}})

e. ({{2{x^2} - 20x + 50} over {3x + 3}}.{{{x^2} - 1} over {4{{left( {x - 5} ight)}^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}}}} ight)) ( = {{24{y^5}.left( { - 21x} ight)} over {7{x^2}.12{y^3}}} = {{2{y^2}.left( { - 3} ight)} over x} =  - {{6{y^2}} over x})

c. (left( { - {{18{y^3}} over {25{x^4}}}} ight).left( { - {{15{x^2}} over {9{y^3}}}} ight)) ( = {{left( { - 18{y^3}} ight).left( { - 15{x^2}} ight)} over {25{x^4}.9{y^3}}} = {{ - 2.left( { - 3} ight)} over {5{x^2}.1}} = {6 over {5{x^2}}})

d. ({{4x + 8} over {{{left( {x - 10} ight)}^3}}}.{{2x - 20} over {{{left( {x + 2} ight)}^2}}})( = {{4left( {x + 2} ight).2left( {x - 10} ight)} over {{{left( {x - 10} ight)}^3}{{left( {x + 2} ight)}^2}}} = {8 over {{{left( {x - 10} ight)}^2}left( {x + 2} ight)}})

e. ({{2{x^2} - 20x + 50} over {3x + 3}}.{{{x^2} - 1} over {4{{left( {x - 5} ight)}^3}}})( = {{2left( {{x^2} - 10x + 25} ight)left( {x + 1} ight)left( {x - 1} ight)} over {3left( {x + 1} ight).4{{left( {x - 5} ight)}^3}}})

( = {{{{left( {x - 5} ight)}^2}left( {x - 1} ight)} over {6{{left( {x - 5} ight)}^3}}} = {{x - 1} over {6left( {x - 5} ight)}})