Câu 3.2 trang 27 SBT Toán 8 tập 1: Rút gọn phân thức:...

Rút gọn phân thức:

Q( = {{{x^{10}} – {x^8} – {x^7} + {x^6} + {x^5} + {x^4} – {x^3} – {x^2} + 1} over {{x^{30}} + {x^{24}} + {x^{18}} + {x^{12}} + {x^6} + 1}})

Giải:

Q( = {{{x^{10}} – {x^8} – {x^7} + {x^6} + {x^5} + {x^4} – {x^3} – {x^2} + 1} over {{x^{30}} + {x^{24}} + {x^{18}} + {x^{12}} + {x^6} + 1}})

(eqalign{  &  = {{left( {{x^{10}} – {x^8} + {x^6}} ight) – left( {{x^7} – {x^5} + {x^3}} ight) + left( {{x^4} – {x^2} + 1} ight)} over {left( {{x^{30}} + {x^{24}} + {x^{18}}} ight) + left( {{x^{12}} + {x^6} + 1} ight)}}  cr  &  = {{{x^6}left( {{x^4} – {x^2} + 1} ight) – {x^3}left( {{x^4} – {x^2} + 1} ight) + left( {{x^4} – {x^2} + 1} ight)} over {{x^{18}}left( {{x^{12}} + {x^6} + 1} ight) + left( {{x^{12}} + {x^6} + 1} ight)}}  cr  &  = {{left( {{x^4} – {x^2} + 1} ight)left( {{x^6} – {x^3} + 1} ight)} over {left( {{x^{12}} + {x^6} + 1} ight)left( {{x^{18}} + 1} ight)}} = {{left( {{x^4} – {x^2} + 1} ight)left( {{x^6} – {x^3} + 1} ight)} over {left( {{x^{12}} + 2{x^6} + 1 – {x^6}} ight)left[ {{{left( {{x^6}} ight)}^3} + 1} ight]}}  cr  &  = {{left( {{x^4} – {x^2} + 1} ight)left( {{x^6} – {x^3} + 1} ight)} over {left[ {{{left( {{x^6} + 1} ight)}^2} – {{left( {{x^3}} ight)}^2}} ight]left( {{x^6} + 1} ight)left( {{x^{12}} – {x^6} + 1} ight)}}  cr  &  = {{left( {{x^4} – {x^2} + 1} ight)left( {{x^6} – {x^3} + 1} ight)} over {left( {{x^6} + {x^3} + 1} ight)left( {{x^6} + 1 – {x^3}} ight)left( {{x^6} + 1} ight)left( {{x^{12}} – {x^6} + 1} ight)}}  cr  &  = {{{x^4} – {x^2} + 1} over {left( {{x^6} + {x^3} + 1} ight)left( {{x^2} + 1} ight)left( {{x^4} – {x^2} + 1} ight)left( {{x^{12}} – {x^6} + 1} ight)}}  cr  &  = {1 over {left( {{x^6} + {x^3} + 1} ight)left( {{x^2} + 1} ight)left( {{x^{12}} – {x^6} + 1} ight)}} cr} )