http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/129437/toan-10
Đặt u=357;v=537" role="present
at
ion" style="font-size: 13.696px; disp
lay:
inline; word-spacing: 0px; position: relat
ive;">u=35−−√7;v=53−−√7u=357;v=537Ta Có u+v=a,uv=1" ro
le="presenta
tion
" style="font-size: 13.
696px; display: inline; position: relative;">u+v=a,uv=1u+v=a,uv=1Dễ dàng cm đc (nh
ác
viết
^^! thông cảm)u2+v2=a
2−2(1)" role="presentation" style="font-size: 13.696px; display: in
line; position: relative;">u2+v2=a2−2(1)u2+v2=a2−2(1)u3+v3=a3−3a(2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">u3+v3=a3−3a(2)u3+v3=a3−3a(2)Nh
ân (1) và (2) vế theo vế =&a
mp;gt;u5+v5=a5−5a3+5a(3)" role="presentation" style="font-size: 13.
696px; display: inline; position: relativ
e;">=>u5+v5=a5−5a3+5a(3)=>u5+v5=a5−5a3+5a(3)Nhân
(1) với (3) vế theo vế ta có u7+v7=a7−7a5+14a3−7a" ro
le="presentati
on" style="font-
size: 13.696px; displa
y: inline; p
osition: relative;">u7+v7=a7−7a5+14a3−7au7+v7=a7−7a5+14a3−7aTa C
ó :u7+v7=3415" role="presenta
tion" style="font-size: 13px; display: inline; position: relative;">u
7+v7=3415u7+v7=3415=>a7−7a5+14a3−7a=34/15" role="presentation" style="font-
size: 13px; display: inline; positio
n: relati
ve;">=>a7−7a5+14a3−7a=34/1
5=>a7−7a5+14
a3
−7
a=34/
15=>
;x" ro
le="presenta
tion
" style="font-
size: 13px; display: inline; position: relative;">=>x=>x là nghiệm của PT:15a7−10
5a5+210a3−105a−34=0" role="presentation" style="font-size: 13px; display: inline; position: relative;">PT:15a7−105a5+210a3−105a−34=0PT:15a7−105a5+210a3−105a−34=0Vậy suy ra $15ka^7-105kx^5+210kx^3-105kx-34k$ ( k là số nguyên khác 0) là đă thức cần tìm!