Câu 36 trang 34 Sách bài tập (SBT) Toán 8 tập 1

Hãy làm các phép chia sau :

a. ({{7x + 2} over {3x{y^3}}}:{{14x + 4} over {{x^2}y}})

b. ({{8xy} over {3x – 1}}:{{12x{y^3}} over {5 – 15x}})

c. ({{27 – {x^3}} over {5x + 5}}:{{2x – 6} over {3x + 3}})

d. (left( {4{x^2} – 16} right):{{3x + 6} over {7x – 2}})

e. ({{3{x^3} + 3} over {x – 1}}:left( {{x^2} – x + 1} right))

Giải:

a. ({{7x + 2} over {3x{y^3}}}:{{14x + 4} over {{x^2}y}})( = {{7x + 2} over {3x{y^3}}}.{{{x^2}y} over {14x + 4}} = {{left( {7x + 2} right){x^2}y} over {3x{y^3}.2left( {7x + 2} right)}} = {x over {6{y^2}}})

b. ({{8xy} over {3x – 1}}:{{12x{y^3}} over {5 – 15x}})( = {{8xy} over {3x – 1}}.{{5 – 15x} over {12x{y^3}}} = {{8xyleft( {5 – 15x} right)} over {left( {3x – 1} right).12x{y^3}}} = {{ – 10left( {3x – 1} right)} over {3left( {3x – 1} right){y^2}}} = {{10} over {3{y^2}}})

c. ({{27 – {x^3}} over {5x + 5}}:{{2x – 6} over {3x + 3}})( = {{27 – {x^3}} over {5x + 5}}:{{3x + 3} over {2x – 6}} = {{left( {{3^3} – {x^3}} right).3left( {x + 1} right)} over {5left( {x + 1} right).2left( {x – 3} right)}})

( = {{ – 3left( {x – 3} right)left( {{x^2} + 3x + 9} right)} over {10left( {x – 3} right)}} = – {{3left( {{x^2} + 3x + 9} right)} over {10}})

d. (left( {4{x^2} – 16} right):{{3x + 6} over {7x – 2}})

( = left( {4{x^2} – 16} right).{{7x – 2} over {3x + 6}} = {{4left( {x + 2} right)left( {x – 2} right)left( {7x – 2} right)} over {3left( {x + 2} right)}})

( = {{4left( {x – 2} right)left( {7x – 2} right)} over 3})

e. ({{3{x^3} + 3} over {x – 1}}:left( {{x^2} – x + 1} right))( = {{3{x^3} + 3} over {x – 1}}.{1 over {{x^2} – x + 1}} = {{3left( {{x^3} + 1} right)} over {left( {x – 1} right)left( {{x^2} – x + 1} right)}} = {{3left( {x + 1} right)left( {{x^2} – x + 1} right)} over {left( {x – 1} right)left( {{x^2} – x + 1} right)}})

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

Câu 37 trang 34 Sách bài tập (SBT) Toán 8 tập 1

Thực hiện phép tính ( chú ý đến quy tắc đổi dấu)

a. ({{4left( {x + 3} right)} over {3{x^2} – x}}:{{{x^2} + 3x} over {1 – 3x}})

b. ({{4x + 6y} over {x – 1}}:{{4{x^2} + 12xy + 9{y^2}} over {1 – {x^3}}})

Giải:

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

b. ({{4x + 6y} over {x – 1}}:{{4{x^2} + 12xy + 9{y^2}} over {1 – {x^3}}} = )({{4x + 6y} over {x – 1}}.{{1 – {x^3}} over {4{x^2} + 12xy + 9{y^2}}} = {{2left( {2x + 3y} right)left( {1 – x} right)left( {1 + x + {x^2}} right)} over {left( {x – 1} right){{left( {2x + 3y} right)}^2}}})

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

Câu 38 trang 34 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn biểu thức :

a. ({{{x^4} – x{y^3}} over {2xy + {y^2}}}:{{{x^3} + {x^2}y + x{y^2}} over {2x + y}})

b. ({{5{x^2} – 10xy + 5{y^2}} over {2{x^2} – 2xy + 2{y^2}}}:{{8x – 8y} over {10{x^3} + 10{y^3}}})

Giải:

a. ({{{x^4} – x{y^3}} over {2xy + {y^2}}}:{{{x^3} + {x^2}y + x{y^2}} over {2x + y}})( = {{{x^4} – x{y^3}} over {2xy + {y^2}}}.{{2x + y} over {{x^3} + {x^2}y + x{y^2}}} = {{xleft( {{x^3} – {y^3}} right)left( {2x + y} right)} over {yleft( {2x + y} right).xleft( {{x^2} + xy + {y^2}} right)}})

( = {{left( {x – y} right)left( {{x^2} + xy + {y^2}} right)} over {yleft( {{x^2} + xy + {y^2}} right)}} = {{x – y} over y})

b. ({{5{x^2} – 10xy + 5{y^2}} over {2{x^2} – 2xy + 2{y^2}}}:{{8x – 8y} over {10{x^3} + 10{y^3}}})( = {{5{x^2} – 10xy + 5{y^2}} over {2{x^2} – 2xy + 2{y^2}}}.{{10{x^3} + 10{y^3}} over {8x – 8y}} = {{5left( {{x^2} – 2xy + {y^2}} right).10left( {{x^3} + {y^3}} right)} over {2left( {{x^2} – xy + {y^2}} right).8left( {x – y} right)}})

( = {{25{{left( {x – y} right)}^2}left( {x + y} right)left( {{x^2} – xy + {y^2}} right)} over {8left( {{x^2} – xy + {y^2}} right)left( {x – y} right)}} = {{25left( {x – y} right)left( {x + y} right)} over 8})

Câu 39 trang 34 Sách bài tập (SBT) Toán 8 tập 1

Thực hiện phép chia phân thức :

a. ({{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}:{{{x^2} – 4x + 4} over {{x^2} + 3x}})

b. ({{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}:{{{x^2} + 7x + 12} over {{x^2} – 9x + 14}})

Giải:

a. ({{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}:{{{x^2} – 4x + 4} over {{x^2} + 3x}})( = {{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}.{{{x^2} + 3x} over {{x^2} – 4x + 4}})

( = {{left( {{x^2} – 5x + 6} right).xleft( {x + 3} right)} over {left( {{x^2} + 7x + 12} right){{left( {x – 2} right)}^2}}} = {{left( {{x^2} – 2x – 3x + 6} right).xleft( {x + 3} right)} over {left( {{x^2} + 3x + 4x + 12} right){{left( {x – 2} right)}^2}}})

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

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

b. ({{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}:{{{x^2} + 7x + 12} over {{x^2} – 9x + 14}})( = {{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}.{{{x^2} – 9x + 14} over {{x^2} + 7x + 12}})

(eqalign{ & = {{left( {{x^2} + 2x – 3} right)left( {{x^2} – 9x + 14} right)} over {left( {{x^2} + 3x – 10} right)left( {{x^2} + 7x + 12} right)}} = {{left( {{x^2} + 3x – x – 3} right)left( {{x^2} – 7x – 2x + 14} right)} over {left( {{x^2} + 5x – 2x + 10} right)left( {{x^2} + 3x + 4x + 12} right)}} cr & = {{left[ {xleft( {x + 3} right) – left( {x + 3} right)} right]left[ {xleft( {x – 7} right) – 2left( {x – 7} right)} right]} over {left[ {xleft( {x + 5} right) – 2left( {x + 5} right)} right]left[ {xleft( {x + 3} right) + 4left( {x + 3} right)} right]}} cr & = {{left( {x + 3} right)left( {x – 1} right)left( {x – 7} right)left( {x – 2} right)} over {left( {x + 5} right)left( {x – 2} right)left( {x + 3} right)left( {x + 4} right)}} = {{left( {x – 1} right)left( {x – 7} right)} over {left( {x + 5} right)left( {x + 4} right)}} cr} )

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