Câu 21 trang 29 Sách bài tập (SBT) Toán 8 tập 1: Làm tính cộng các phân thức...

Làm tính cộng các phân thức

a. ({{11x + 13} over {3x – 3}} + {{15x + 17} over {4 – 4x}})

b. ({{2x + 1} over {2{x^2} – x}} + {{32{x^2}} over {1 – 4{x^2}}} + {{1 – 2x} over {2{x^2} + x}})

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

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

Giải:

a. ({{11x + 13} over {3x – 3}} + {{15x + 17} over {4 – 4x}})( = {{11x + 13} over {3left( {x – 1} ight)}} + {{ – 15x – 17} over {4left( {x – 1} ight)}})

(eqalign{  &  = {{4left( {11x + 13} ight)} over {12left( {x – 1} ight)}} + {{3left( { – 15x – 17} ight)} over {12left( {x – 1} ight)}} = {{44x + 52 – 45x – 51} over {12left( {x – 1} ight)}} = {{1 – x} over {12left( {x – 1} ight)}}  cr  &  = {{ – left( {x – 1} ight)} over {12left( {x – 1} ight)}} =  – {1 over {12}} cr} )

b. ({{2x + 1} over {2{x^2} – x}} + {{32{x^2}} over {1 – 4{x^2}}} + {{1 – 2x} over {2{x^2} + x}})( = {{2x + 1} over {xleft( {2x – 1} ight)}} + {{ – 32{x^2}} over {left( {2x + 1} ight)left( {2x – 1} ight)}} + {{1 – 2x} over {xleft( {2x + 1} ight)}})

(eqalign{  &  = {{left( {2x + 1} ight)left( {2x + 1} ight)} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}} + {{ – 32{x^2}.x} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}} + {{left( {1 – 2x} ight)left( {2x – 1} ight)} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}}  cr  &  = {{4{x^2} + 4x + 1 – 32{x^3} + 2x – 1 – 4{x^2} + 2x} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}} = {{ – 32{x^3} + 8x} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}}  cr  &  = {{ – 8xleft( {4{x^2} – 1} ight)} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}} = {{ – 8xleft( {2x + 1} ight)left( {2x – 1} ight)} over {xleft( {2x + 1} ight)left( {2x – 1} ight)}} =  – 8 cr} )

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

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

d. ({{{x^4}} over {1 – x}} + {x^3} + {x^2} + x + 1)( = {{{x^4}} over {1 – x}} + {{left( {{x^3} + {x^2} + x + 1} ight)left( {1 – x} ight)} over {1 – x}})

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