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)}})

Câu 30 trang 32 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn biểu thức (chú ý dùng quy tắc đổi dấu để thấy nhân tử chung) :

a. ({{x + 3} over {{x^2} – 4}}.{{8 – 12x + 6{x^2} – {x^3}} over {9x + 27}})

b. ({{6x – 3} over {5{x^2} + x}}.{{25{x^2} + 10x + 1} over {1 – 8{x^3}}})

c. ({{3{x^2} – x} over {{x^2} – 1}}.{{1 – {x^4}} over {{{left( {1 – 3x} right)}^3}}})

Giải:

a. ({{x + 3} over {{x^2} – 4}}.{{8 – 12x + 6{x^2} – {x^3}} over {9x + 27}})({{left( {x + 3} right)left( {8 – 12x + 6{x^2} – {x^3}} right)} over {left( {x + 2} right)left( {x – 2} right).9left( {x + 3} right)}})

( = {{{2^3} – {{3.2}^2}.x + 3.2{x^2} – {x^3}} over {9left( {x + 2} right)left( {x – 2} right)}} = {{{{left( {2 – x} right)}^3}} over { – 9left( {x + 2} right)left( {2 – x} right)}} = – {{{{left( {2 – x} right)}^2}} over {9left( {x + 2} right)}})

b. ({{6x – 3} over {5{x^2} + x}}.{{25{x^2} + 10x + 1} over {1 – 8{x^3}}})( = {{3left( {2x – 1} right){{left( {5x + 1} right)}^2}} over {xleft( {5x + 1} right)left[ {1 – {{left( {2x} right)}^2}} right]}} = {{3left( {2x – 1} right)left( {5x + 1} right)} over {xleft( {1 – 2x} right)left( {1 + 2x + 4{x^2}} right)}})

( = – {{3left( {2x – 1} right)left( {5x + 1} right)} over {xleft( {2x – 1} right)left( {1 + 2x + 4{x^2}} right)}} = – {{3left( {5x + 1} right)} over {xleft( {1 + 2x + 4{x^2}} right)}})

c. ({{3{x^2} – x} over {{x^2} – 1}}.{{1 – {x^4}} over {{{left( {1 – 3x} right)}^3}}})( = {{xleft( {3x – 1} right)left( {1 – {x^4}} right)} over {left( {{x^2} – 1} right){{left( {1 – 3x} right)}^3}}} = {{xleft( {3x – 1} right)left( {{x^2} – 1} right)left( {{x^2} + 1} right)} over {left( {{x^2} – 1} right){{left( {3x – 1} right)}^3}}})

( = {{xleft( {{x^2} + 1} right)} over {{{left( {3x – 1} right)}^2}}})

Câu 31 trang 32 Sách bài tập (SBT) Toán 8 tập 1

Phân tích các tử thức và các mẫu thức (nếu cần thì dùng phương pháp thêm và bớt cùng một số hạng hoặc tách một số hạng thành hai số hạng) rồi rút gọn biểu thức :

a. ({{x – 2} over {x + 1}}.{{{x^2} – 2x – 3} over {{x^2} – 5x + 6}})

b. ({{x + 1} over {{x^2} – 2x – 8}}.{{4 – x} over {{x^2} + x}})

c. ({{x + 2} over {4x + 24}}.{{{x^2} – 36} over {{x^2} + x – 2}})

Giải:

a. ({{x – 2} over {x + 1}}.{{{x^2} – 2x – 3} over {{x^2} – 5x + 6}})( = {{left( {x – 2} right)left( {{x^2} – 2x – 3} right)} over {left( {x + 1} right)left( {{x^2} – 5x + 6} right)}} = {{left( {x – 2} right)left( {{x^2} – 3x + x – 3} right)} over {left( {x + 1} right)left( {{x^2} – 2x – 3x + 6} right)}})

( = {{left( {x – 2} right)left[ {xleft( {x – 3} right) + left( {x – 3} right)} right]} over {left( {x + 1} right)left[ {xleft( {x – 2} right) – 3left( {x – 2} right)} right]}} = {{left( {x – 2} right)left( {x – 3} right)left( {x + 1} right)} over {left( {x + 1} right)left( {x – 2} right)left( {x – 3} right)}} = 1)

b. ({{x + 1} over {{x^2} – 2x – 8}}.{{4 – x} over {{x^2} + x}})( = {{left( {x + 1} right)left( {4 – x} right)} over {left( {{x^2} – 2x – 8} right)xleft( {x + 1} right)}} = {{4 – x} over {left( {{x^2} – 4x + 2x – 8} right)x}})

( = {{4 – x} over {left[ {xleft( {x – 4} right) + 2left( {x – 4} right)} right]x}} = {{4 – x} over {xleft( {x – 4} right)left( {x + 2} right)}} = – {{x – 4} over {xleft( {x – 4} right)left( {x + 2} right)}} = – {1 over {xleft( {x + 2} right)}})

c. ({{x + 2} over {4x + 24}}.{{{x^2} – 36} over {{x^2} + x – 2}})({{left( {x + 2} right)left( {x + 6} right)left( {x – 6} right)} over {4left( {x + 6} right)left( {{x^2} + x – 2} right)}} = {{left( {x + 2} right)left( {x – 6} right)} over {4left( {{x^2} + 2x – x – 2} right)}})

( = {{left( {x + 2} right)left( {x – 6} right)} over {4left[ {xleft( {x + 2} right) – left( {x – 2} right)} right]}} = {{left( {x + 2} right)left( {x – 6} right)} over {4left( {x + 2} right)left( {x – 1} right)}} = {{x – 6} over {4left( {x – 1} right)}})

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