Câu 19 trang 29 Sách bài tập Toán 8 tập 1: Dùng quy tắc đổi dấu để tìm mẫu thức chung rồi thực hiện phép...

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} ight)left( {x – 2} ight)}})

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

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

(eqalign{  &  = {{{{left( {x – 3} ight)}^2}} over {{{left( {x + 3} ight)}^2}{{left( {x – 3} ight)}^2}}} + {{ – {{left( {x + 3} ight)}^2}} over {{{left( {x + 3} ight)}^2}{{left( {x – 3} ight)}^2}}} + {{xleft( {x + 3} ight)left( {x – 3} ight)} over {{{left( {x + 3} ight)}^2}{{left( {x – 3} ight)}^2}}}  cr  &  = {{{x^2} – 6x + 9 – {x^2} – 6x – 9 + {x^3} – 9x} over {{{left( {x + 3} ight)}^2}{{left( {x – 3} ight)}^2}}} = {{{x^3} – 21x} over {{{left( {x + 3} ight)}^2}{{left( {x – 3} ight)}^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} ight)left( {{x^2} + x + 1} ight)}} + {2 over {{x^2} + x + 1}} + {{ – 1} over {x – 1}})

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

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