Giải bài tập trang 39, 40 bài ôn tập Chương II – Phân thức đại số Sách bài tập (SBT) Toán 8 tập 1. Câu 58: Thực hiện các phép tính …
Câu 58 trang 39 Sách bài tập (SBT) Toán 8 tập 1
Thực hiện các phép tính :
a. (left( {{9 over {{x^3} – 9x}} + {1 over {x + 3}}} right):left( {{{x – 3} over {{x^2} + 3x}} – {x over {3x + 9}}} right))
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b. (left( {{2 over {x – 2}} – {2 over {x + 2}}} right).{{{x^2} + 4x + 4} over 8})
c. (left( {{{3x} over {1 – 3x}} + {{2x} over {3x + 1}}} right):{{6{x^2} + 10x} over {1 – 6x + 9{x^2}}})
d. (left( {{x over {{x^2} – 25}} – {{x – 5} over {{x^2} + 5x}}} right):{{2x – 5} over {{x^2} + 5x}} + {x over {5 – x}})
e. (left( {{{{x^2} + xy} over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y over {{x^2} + {y^2}}}} right):left( {{1 over {x – y}} – {{2xy} over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} right))
Giải:
a. (left( {{9 over {{x^3} – 9x}} + {1 over {x + 3}}} right):left( {{{x – 3} over {{x^2} + 3x}} – {x over {3x + 9}}} right))
(eqalign{ & = left[ {{9 over {xleft( {x + 3} right)left( {x – 3} right)}} + {1 over {x + 3}}} right]:left[ {{{x – 3} over {xleft( {x + 3} right)}} – {x over {3left( {x + 3} right)}}} right] cr & = {{9 + xleft( {x – 3} right)} over {xleft( {x + 3} right)left( {x – 3} right)}}:{{3left( {x – 3} right) – {x^2}} over {3xleft( {x + 3} right)}} = {{{x^2} – 3x + 9} over {xleft( {x + 3} right)left( {x – 3} right)}}.{{3xleft( {x + 3} right)} over {3x – 9 – {x^2}}} cr & = {{3left( {{x^2} – 3x + 9} right)} over {left( {3 – x} right)left( {{x^2} – 3x + 9} right)}} = {3 over {3 – x}} cr} )
b. (left( {{2 over {x – 2}} – {2 over {x + 2}}} right).{{{x^2} + 4x + 4} over 8})( = {{2left( {x + 2} right) – 2left( {x – 2} right)} over {left( {x – 2} right)left( {x + 2} right)}}.{{{{left( {x + 2} right)}^2}} over 8})
( = {{2x + 4 – 2x + 4} over {left( {x – 2} right)left( {x + 2} right)}}.{{{{left( {x + 2} right)}^2}} over 8} = {8 over {left( {x – 2} right)left( {x + 2} right)}}.{{{{left( {x + 2} right)}^2}} over 8} = {{x + 2} over {x – 2}})
c. (left( {{{3x} over {1 – 3x}} + {{2x} over {3x + 1}}} right):{{6{x^2} + 10x} over {1 – 6x + 9{x^2}}})( = {{3xleft( {3x + 1} right) + 2xleft( {1 – 3x} right)} over {left( {1 – 3x} right)left( {1 + 3x} right)}}:{{2xleft( {3x + 5} right)} over {{{left( {1 – 3x} right)}^2}}})
(eqalign{ & = {{9{x^2} + 3x + 2x – 6{x^2}} over {left( {1 – 3x} right)left( {1 + 3x} right)}}.{{{{left( {1 – 3x} right)}^2}} over {2xleft( {3x + 5} right)}} = {{xleft( {3x + 5} right)} over {left( {1 – 3x} right)left( {1 + 3x} right)}}.{{{{left( {1 – 3x} right)}^2}} over {2xleft( {3x + 5} right)}} cr & = {{1 – 3x} over {2left( {1 + 3x} right)}} cr} )
d. (left( {{x over {{x^2} – 25}} – {{x – 5} over {{x^2} + 5x}}} right):{{2x – 5} over {{x^2} + 5x}} + {x over {5 – x}})
(eqalign{ & = left[ {{x over {left( {x + 5} right)left( {x – 5} right)}} – {{x – 5} over {xleft( {x + 5} right)}}} right]:{{2x – 5} over {xleft( {x + 5} right)}} + {x over {5 – x}} cr & = {{{x^2} – {{left( {x – 5} right)}^2}} over {xleft( {x + 5} right)left( {x – 5} right)}}.{{xleft( {x + 5} right)} over {2x – 5}} + {x over {5 – x}} cr & = {{{x^2} – {x^2} + 10x – 25} over {left( {x – 5} right)left( {2x – 5} right)}} + {x over {5 – x}} = {{5left( {2x – 5} right)} over {left( {x – 5} right)left( {2x – 5} right)}} – {x over {x – 5}} cr & = {5 over {x – 5}} – {x over {x – 5}} = {{5 – x} over {x – 5}} = {{ – left( {x – 5} right)} over {x – 5}} = – 1 cr} )
e. (left( {{{{x^2} + xy} over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y over {{x^2} + {y^2}}}} right):left( {{1 over {x – y}} – {{2xy} over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} right))
(eqalign{ & = left[ {{{{x^2} + xy} over {left( {{x^2} + {y^2}} right)left( {x + y} right)}} + {y over {{x^2} + {y^2}}}} right]:left[ {{1 over {x – y}} – {{2xy} over {left( {{x^2} + {y^2}} right)left( {x – y} right)}}} right] cr & = {{{x^2} + xy + yleft( {x + y} right)} over {left( {{x^2} + {y^2}} right)left( {x + y} right)}}:{{{x^2} + {y^2} – 2xy} over {left( {{x^2} + {y^2}} right)left( {x – y} right)}} cr & = {{{x^2} + xy + xy + {y^2}} over {left( {{x^2} + {y^2}} right)left( {x + y} right)}}.{{left( {{x^2} + {y^2}} right)left( {x – y} right)} over {{{left( {x – y} right)}^2}}} cr & = {{{{left( {x + y} right)}^2}} over {left( {{x^2} + {y^2}} right)left( {x + y} right)}}.{{left( {{x^2} + {y^2}} right)left( {x – y} right)} over {{{left( {x – y} right)}^2}}} = {{x + y} over {x – y}} cr} )
Câu 59 trang 40 Sách bài tập (SBT) Toán 8 tập 1
Chứng minh đẳng thức :
a. (left( {{{{x^2} – 2x} over {2{x^2} + 8}} – {{2{x^2}} over {8 – 4x + 2{x^2} – {x^3}}}} right)left( {1 – {1 over x} – {2 over {{x^2}}}} right) = {{x + 1} over {2x}})
b. (left[ {{2 over {3x}} – {2 over {x + 1}}.left( {{{x + 1} over {3x}} – x – 1} right)} right]:{{x – 1} over x} = {{2x} over {x – 1}})
c. (left[ {{2 over {{{left( {x + 1} right)}^3}}}.left( {{1 over x} + 1} right) + {1 over {{x^2} + 2x + 1}}.left( {{1 over {{x^2}}} + 1} right)} right]:{{x – 1} over {{x^3}}} = {x over {x – 1}})
Giải:
a. Biến đổi vế trái :
(left( {{{{x^2} – 2x} over {2{x^2} + 8}} – {{2{x^2}} over {8 – 4x + 2{x^2} – {x^3}}}} right)left( {1 – {1 over x} – {2 over {{x^2}}}} right))
(eqalign{ & = left[ {{{{x^2} – 2x} over {2left( {{x^2} + 4} right)}} – {{2{x^2}} over {4left( {2 – x} right) + {x^2}left( {2 – x} right)}}} right]{{{x^2} – x – 2} over {{x^2}}} cr & = left[ {{{{x^2} – 2x} over {2left( {{x^2} + 4} right)}} – {{2{x^2}} over {left( {2 – x} right)left( {4 + {x^2}} right)}}} right]{{{x^2} – x – 2} over {{x^2}}} cr & = {{left( {{x^2} – 2x} right)left( {2 – x} right) – 4{x^2}} over {2left( {2 – x} right)left( {{x^2} + 4} right)}}.{{{x^2} – x – 2} over {{x^2}}} cr & = {{2{x^2} – {x^3} – 4x + 2{x^2} – 4{x^2}} over {2left( {2 – x} right)left( {{x^2} + 4} right)}}.{{{x^2} – 2x + x – 2} over {{x^2}}} cr & = {{ – xleft( {{x^2} + 4} right)} over {2left( {2 – x} right)left( {{x^2} + 4} right)}}.{{xleft( {x – 2} right) + left( {x – 2} right)} over {{x^2}}} cr & = {{xleft( {{x^2} + 4} right)} over {2left( {x – 2} right)left( {{x^2} + 4} right)}}.{{left( {x – 2} right)left( {x + 1} right)} over {{x^2}}} = {{x + 1} over {2x}} cr} )
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
b. Biến đổi vế trái:
(eqalign{ & left[ {{2 over {3x}} – {2 over {x + 1}}.left( {{{x + 1} over {3x}} – x – 1} right)} right]:{{x – 1} over x} cr & = left[ {{2 over {3x}} – {2 over {x + 1}}.{{x + 1 – 3xleft( {x + 1} right)} over {3x}}} right].{x over {x – 1}} cr & = left[ {{2 over {3x}} – {2 over {x + 1}}.{{left( {x + 1} right)left( {1 – 3x} right)} over {3x}}} right].{x over {x – 1}} cr & = left[ {{2 over {3x}} – {{2left( {1 – 3x} right)} over {3x}}} right].{x over {x – 1}} = {{2 – 2 + 6x} over {3x}}.{x over {x – 1}} = 2.{x over {x – 1}} = {{2x} over {x – 1}} cr} )
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
c. Biến đổi vế trái :
(eqalign{ & left[ {{2 over {{{left( {x + 1} right)}^3}}}.left( {{1 over x} + 1} right) + {1 over {{x^2} + 2x + 1}}.left( {{1 over {{x^2}}} + 1} right)} right]:{{x – 1} over {{x^3}}} cr & = left[ {{2 over {{{left( {x + 1} right)}^3}}}.{{x + 1} over x} + {1 over {{{left( {x + 1} right)}^2}}}.{{{x^2} + 1} over {{x^2}}}} right].{{{x^3}} over {x – 1}} cr & = left[ {{2 over {x{{left( {x + 1} right)}^2}}} + {{{x^2} + 1} over {{x^2}{{left( {x + 1} right)}^2}}}} right].{{{x^3}} over {x – 1}} = {{2x + {x^2} + 1} over {{x^2}{{left( {x + 1} right)}^2}}}.{{{x^3}} over {x – 1}} cr & = {{{{left( {x + 1} right)}^2}} over {{x^2}{{left( {x + 1} right)}^2}}}.{{{x^3}} over {x – 1}} = {x over {x – 1}} cr} )
Vế trái bằng vế phải, vậy đẳng thức được chứng minh.
Câu 60 trang 40 Sách bài tập (SBT) Toán 8 tập 1
Biến đổi các biểu thức hữu tỉ thành phân thức :
a. ({{{x over {x – 1}} – {{x + 1} over x}} over {{x over {x + 1}} – {{x – 1} over x}}})
b. ({{{5 over 4} – {5 over {x + 1}}} over {{{9 – {x^2}} over {{x^2} + 2x + 1}}}})
Giải:
a. ({{{x over {x – 1}} – {{x + 1} over x}} over {{x over {x + 1}} – {{x – 1} over x}}})( = left( {{x over {x – 1}} – {{x + 1} over x}} right):left( {{x over {x + 1}} – {{x – 1} over x}} right))
( = {{{x^2} – left( {x + 1} right)left( {x – 1} right)} over {xleft( {x – 1} right)}}:{{{x^2} – left( {x – 1} right)left( {x + 1} right)} over {xleft( {x + 1} right)}} = {1 over {xleft( {x – 1} right)}}.{{xleft( {x + 1} right)} over 1} = {{x + 1} over {x – 1}})
b. ({{{5 over 4} – {5 over {x + 1}}} over {{{9 – {x^2}} over {{x^2} + 2x + 1}}}})( = left( {{5 over 4} – {5 over {x + 1}}} right):left( {{{9 – {x^2}} over {{x^2} + 2x + 1}}} right) = {{5left( {x + 1} right) – 20} over {4left( {x + 1} right)}}:{{left( {3 + x} right)left( {3 – x} right)} over {{{left( {x + 1} right)}^2}}})
( = {{5left( {x – 3} right)} over {4left( {x + 1} right)}}.{{{{left( {x + 1} right)}^2}} over {left( {3 + x} right)left( {3 – x} right)}} = {{ – 5left( {3 – x} right)left( {x + 1} right)} over {4left( {3 + x} right)left( {3 – x} right)}} = {{ – 5left( {x + 1} right)} over {4left( {3 + x} right)}})
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