Giải bài 17, 18, 19 trang 28, 29 SBT Toán 8 tập 1

Giải bài tập trang 28, 29 bài 5 phép cộng các phân thức đại số Sách bài tập (SBT) Toán 8 tập 1. Câu 17: Cộng các phân thức cùng mẫu thức…

Câu 17 trang 28 Sách bài tập (SBT) Toán 8 tập 1

Cộng các phân thức cùng mẫu thức

a. ({{1 – 2x} over {6{x^3}y}} + {{3 + 2y} over {6{x^3}y}} + {{2x – 4} over {6{x^3}y}})

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b. ({{{x^2} – 2} over {x{{left( {x – 1} right)}^2}}} + {{2 – x} over {x{{left( {x – 1} right)}^2}}})

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

d. ({{{x^2} + 38x + 4} over {2{x^2} + 17x + 1}} + {{3{x^2} – 4x – 2} over {2{x^2} + 17x + 1}})

Giải:

a. ({{1 – 2x} over {6{x^3}y}} + {{3 + 2y} over {6{x^3}y}} + {{2x – 4} over {6{x^3}y}}) ( = {{1 – 2x + 3 + 2y + 2x – 4} over {6{x^3}y}} = {{2y} over {6{x^3}y}} = {1 over {3{x^3}}})

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

c. ({{3x + 1} over {{x^2} – 3x + 1}} + {{{x^2} – 6x} over {{x^2} – 3x + 1}}) ( = {{3x + 1 + {x^2} – 6x} over {{x^2} – 3x + 1}} = {{{x^2} – 3x + 1} over {{x^2} – 3x + 1}} = 1)

d. ({{{x^2} + 38x + 4} over {2{x^2} + 17x + 1}} + {{3{x^2} – 4x – 2} over {2{x^2} + 17x + 1}}) ( = {{{x^2} + 38x + 4 + 3{x^2} – 4x – 2} over {2{x^2} + 17x + 1}} = {{4{x^2} + 34x + 2} over {2{x^2} + 17x + 1}} = {{2left( {2{x^2} + 17x + 1} right)} over {2{x^2} + 17x + 1}} = 2)

Câu 18 trang 28 Sách bài tập (SBT) Toán 8 tập 1

Cộng các phân thức khác mẫu thức:

a. ({5 over {6{x^2}y}} + {7 over {12x{y^2}}} + {{11} over {18xy}})

b. ({{4x + 2} over {15{x^3}y}} + {{5y – 3} over {9{x^2}y}} + {{x + 1} over {5x{y^3}}})

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

d. ({{{x^3} + 2x} over {{x^3} + 1}} + {{2x} over {{x^2} – x + 1}} + {1 over {x + 1}})

Giải:

a. ({5 over {6{x^2}y}} + {7 over {12x{y^2}}} + {{11} over {18xy}})( = {{30y} over {36{x^2}{y^2}}} + {{21x} over {36{x^2}{y^2}}} + {{22xy} over {36{x^2}{y^2}}} = {{30y + 21x + 22xy} over {36{x^2}{y^2}}})

b. ({{4x + 2} over {15{x^3}y}} + {{5y – 3} over {9{x^2}y}} + {{x + 1} over {5x{y^3}}})(eqalign{  &  = {{3{y^2}left( {4x + 2} right)} over {45{x^3}{y^3}}} + {{5x{y^2}left( {5y – 3} right)} over {45{x^3}{y^3}}} + {{9{x^2}left( {x + 1} right)} over {45{x^3}{y^3}}}  cr  &  = {{12x{y^2} + 6{y^2} + 25x{y^3} – 15x{y^2} + 9{x^3} + 9{x^2}} over {45{x^3}{y^3}}} = {{6{y^2} + 25x{y^3} – 3x{y^2} + 9{x^3} + 9{x^2}} over {45{x^3}{y^3}}} cr} )

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

(eqalign{  &  = {{3left( {2x – 1} right)} over {2xleft( {2x – 1} right)}} + {{2xleft( {3x – 3} right)} over {2xleft( {2x – 1} right)}} + {{2{x^2} + 1} over {2xleft( {2x – 1} right)}} = {{6x – 3 + 6{x^2} – 6x + 2{x^2} + 1} over {2xleft( {2x – 1} right)}}  cr  &  = {{8{x^2} – 2} over {2xleft( {2x – 1} right)}} = {{2left( {4{x^2} – 1} right)} over {2xleft( {2x – 1} right)}} = {{left( {2x + 1} right)left( {2x – 1} right)} over {xleft( {2x – 1} right)}} = {{2x + 1} over x} cr} )

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

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

Câu 19 trang 29 Sách bài tập (SBT) Toán 8 tập 1

Dùng quy tắc đổi dấu để tìm mẫu thức chung rồi thực hiện phép cộng:

a. ({4 over {x + 2}} + {2 over {x – 2}} + {{5x – 6} over {4 – {x^2}}})

b. ({{1 – 3x} over {2x}} + {{3x – 2} over {2x – 1}} + {{3x – 2} over {2x – 4{x^2}}})

c. ({1 over {{x^2} + 6x + 9}} + {1 over {6x – {x^2} – 9}} + {x over {{x^2} – 9}})

d. ({{{x^2} + 2} over {{x^3} – 1}} + {2 over {{x^2} + x + 1}} + {1 over {1 – x}})

e. ({x over {x – 2y}} + {x over {x + 2y}} + {{4xy} over {4{y^2} – {x^2}}})

Giải:

a. ({4 over {x + 2}} + {2 over {x – 2}} + {{5x – 6} over {4 – {x^2}}}) ( = {4 over {x + 2}} + {2 over {x – 2}} + {{6 – 5x} over {left( {x + 2} right)left( {x – 2} right)}})

(eqalign{  &  = {{4left( {x – 2} right)} over {left( {x + 2} right)left( {x – 2} right)}} + {{2left( {x + 2} right)} over {left( {x + 2} right)left( {x – 2} right)}} + {{6 – 5x} over {left( {x + 2} right)left( {x – 2} right)}} = {{4x – 8 + 2x + 4 + 6 – 5x} over {left( {x + 2} right)left( {x – 2} right)}}  cr  &  = {{x + 2} over {left( {x + 2} right)left( {x – 2} right)}} = {1 over {x – 2}} cr} )

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

(eqalign{  &  = {{left( {1 – 3x} right)left( {2x – 1} right)} over {2xleft( {2x – 1} right)}} + {{left( {3x – 2} right).2x} over {2xleft( {2x – 1} right)}} + {{2 – 3x} over {2xleft( {2x – 1} right)}}  cr  &  = {{2x – 1 – 6{x^2} + 3x + 6{x^2} – 4x + 2 – 3x} over {2xleft( {2x – 1} right)}} = {{1 – 2x} over {2xleft( {2x – 1} right)}} = {{ – left( {2x – 1} right)} over {2xleft( {2x – 1} right)}} = {{ – 1} over {2x}} cr} )

c. ({1 over {{x^2} + 6x + 9}} + {1 over {6x – {x^2} – 9}} + {x over {{x^2} – 9}})( = {1 over {{{left( {x + 3} right)}^2}}} + {{ – 1} over {{{left( {x – 3} right)}^2}}} + {x over {left( {x + 3} right)left( {x – 3} right)}})

(eqalign{  &  = {{{{left( {x – 3} right)}^2}} over {{{left( {x + 3} right)}^2}{{left( {x – 3} right)}^2}}} + {{ – {{left( {x + 3} right)}^2}} over {{{left( {x + 3} right)}^2}{{left( {x – 3} right)}^2}}} + {{xleft( {x + 3} right)left( {x – 3} right)} over {{{left( {x + 3} right)}^2}{{left( {x – 3} right)}^2}}}  cr  &  = {{{x^2} – 6x + 9 – {x^2} – 6x – 9 + {x^3} – 9x} over {{{left( {x + 3} right)}^2}{{left( {x – 3} right)}^2}}} = {{{x^3} – 21x} over {{{left( {x + 3} right)}^2}{{left( {x – 3} right)}^2}}} cr} )

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

(eqalign{  &  = {{{x^2} + 2} over {left( {x – 1} right)left( {{x^2} + x + 1} right)}} + {{2left( {x – 1} right)} over {left( {x – 1} right)left( {{x^2} + x + 1} right)}} + {{ – left( {{x^2} + x + 1} right)} over {left( {x – 1} right)left( {{x^2} + x + 1} right)}}  cr  &  = {{{x^2} + 2 + 2x – 2 – {x^2} – x – 1} over {left( {x – 1} right)left( {{x^2} + x + 1} right)}} = {{x – 1} over {left( {x – 1} right)left( {{x^2} + x + 1} right)}} = {1 over {{x^2} + x + 1}} cr} )

e. ({x over {x – 2y}} + {x over {x + 2y}} + {{4xy} over {4{y^2} – {x^2}}})( = {x over {x – 2y}} + {x over {x + 2y}} + {{ – 4xy} over {left( {x + 2y} right)left( {x – 2y} right)}})

(eqalign{  &  = {{xleft( {x + 2y} right)} over {left( {x – 2y} right)left( {x + 2y} right)}} + {{xleft( {x – 2y} right)} over {left( {x – 2y} right)left( {x + 2y} right)}} + {{ – 4xy} over {left( {x – 2y} right)left( {x + 2y} right)}}  cr  &  = {{{x^2} + 2xy + {x^2} – 2xy – 4xy} over {left( {x – 2y} right)left( {x + 2y} right)}} = {{2{x^2} – 4xy} over {left( {x – 2y} right)left( {x + 2y} right)}} = {{2xleft( {x – 2y} right)} over {left( {x – 2y} right)left( {x + 2y} right)}}  cr  &  = {{2x} over {x + 2y}} cr} )

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