Câu 26 trang 31 Sách bài tập (SBT) Toán 8 tập 1

Rút gọn biểu thức :

Rút gọn biểu thức :

a. ({{3{x^2} + 5x + 1} over {{x^3} - 1}} - {{1 - x} over {{x^2} + x + 1}} - {3 over {x - 1}})

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

c. ({7 over x} - {x over {x + 6}} + {{36} over {{x^2} + 6x}})

Giải:

a. ({{3{x^2} + 5x + 1} over {{x^3} - 1}} - {{1 - x} over {{x^2} + x + 1}} - {3 over {x - 1}})

(eqalign{  &  = {{3{x^2} + 5x + 1} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}} + {{x - 1} over {{x^2} + x + 1}} + {{ - 3} over {x - 1}}  cr  &  = {{3{x^2} + 5x + 1} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}} + {{{{left( {x - 1} ight)}^2}} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}} + {{ - 3left( {{x^2} + x + 1} ight)} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}}  cr  &  = {{3{x^2} + 5x + 1 + {x^2} - 2x + 1 - 3{x^2} - 3x - 3} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}} = {{{x^2} - 1} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}}  cr  &  = {{left( {x + 1} ight)left( {x - 1} ight)} over {left( {x - 1} ight)left( {{x^2} + x + 1} ight)}} = {{x + 1} over {{x^2} + x + 1}} cr} )

b. ({1 over {{x^2} - x + 1}} + 1 - {{{x^2} + 2} over {{x^3} + 1}})( = {1 over {{x^2} - x + 1}} + 1 + {{ - left( {{x^2} + 2} ight)} over {left( {x - 1} ight)left( {{x^2} - x + 1} ight)}})

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

c. ({7 over x} - {x over {x + 6}} + {{36} over {{x^2} + 6x}})( = {7 over x} + {{ - x} over {x + 6}} + {{36} over {{x^2} + 6x}} = {{7left( {x + 6} ight)} over {xleft( {x + 6} ight)}} + {{ - {x^2}} over {xleft( {x + 6} ight)}} + {{36} over {xleft( {x + 6} ight)}})

(eqalign{  &  = {{7x + 42 - {x^2} + 36} over {xleft( {x + 6} ight)}} = {{7x - {x^2} + 78} over {xleft( {x + 6} ight)}} = {{13x + 78 - 6x - {x^2}} over {xleft( {x + 6} ight)}}  cr  &  = {{13left( {x + 6} ight) - xleft( {x + 6} ight)} over {xleft( {x + 6} ight)}} = {{left( {x + 6} ight)left( {13 - x} ight)} over {xleft( {x + 6} ight)}} = {{13 - x} over x} cr} )