Câu 26 trang 31 Sách bài tập 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} )