http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/131652/toan-10
Ko mất tính
t
ổng quát
, giả sử x2&
;#x2265;x1" role="presentation" style="font-size:
13.696px; display: inline; word-spacing: 0px; posit
io
n: rela
tive;">x2≥x1x2≥x1Ta có : x4x3=x3x2=x2x1≥1" role="presen
tation" style="font-size: 13.
696px; display: inline; position: relative;">x4x3=x3x2=x2x1≥1x4x3=x3x2=x2x1≥1⇒x4≥x3≥x2≥x1" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒x4≥x3≥x2≥x1⇒x4≥x3≥x2≥x1Ph
ương trình x2−3x+m=0" ro
le="presentation" style="font-size: 13.696px; display: inline; position: relative;">x2−3x+m=0x2−3x+m=0 c
ó 2" role="presenta
ti
on
" style="font-size: 13.696px; display: inline; position: relative;">22 ngh
iệm x2&a
mp;#x2265;x1làx2=3+Δ12;x2=3−Δ12" role="presentation" style="font-size: 13.
696px; display: inline; position: relativ
e;">x2≥x1làx2=3+Δ1−−−√2;x2=3−Δ1−−−√2x2≥x1làx2=3+Δ12;x2=3−Δ12 với Δ1=9−4m≥0" role="presen
tation" style="font-size: 13.696px; display: inline; position: relative;">Δ1=9−4m≥0Δ1=9−4m≥0 hay m≤94" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">m≤94m≤94Phương trình x2−12x+n" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x2−12x+nx2−12x+n có 2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">22 nghiệm x4≥x3làx4=6+Δ2;x3=6−Δ2" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">x4≥x3làx4=6+Δ2−−−√;x3=6−Δ2−−−√x4≥x3làx4=6+Δ2;x3=6−Δ2 với đk Δ2=36−n≥0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">Δ2=36−n≥0Δ2=36−n≥0 hay n≤36" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">n≤36n≤36∗x2x1=x4x3⇒3+Δ13−Δ1=6+Δ26−Δ2" ro
le="presentati
on" style="font-
size: 13.696px; display: inline; position: relative;">∗x2x1=x4x3⇒3+Δ1−−−√3−Δ1−−−√=6+Δ2−−−√6−Δ2−−−√∗x2x1=x4x3⇒3+Δ13−Δ1=6+Δ26−Δ2Nhân chéo rút gọn đc 2Δ1=Δ2(1)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">2Δ1−−−√=Δ2−−−√(1)2Δ1=Δ2(1) ∗x2x1=x3x2ȝ
2;x22=x1x3&a
mp
;#x21D2;(3+Δ12)2=(3−Δ12)(6−Δ2)(2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">∗x2x1=x3x2⇒x22=x1x3⇒(3+Δ1−−−√2)2=(3−Δ1−−−√2)(6−Δ2−−−√)(2)∗x2x1=x3x2⇒x22=x1x3⇒(3+Δ12)2=(3−Δ12)(6−Δ2)(2)Thế (1)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(1)(1) vào (2)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">(2)(2), ta
có (3+Δ12)2=(3−Δ12)(6−2Δ1)" role="presentation" style="font-
size: 13.696px; display: inline; po
si
tion: relative;">(3+Δ1
−−−√2)2=(3
−Δ1
−−−√2)(6
−2
Δ1−−−√)(3+Δ12)2=(3−Δ12)(6−2Δ1)⇒(3+Δ12)2=(3−Δ1)2" role="presentation" style="font-size: 13.696px; display: inline; posit
io
n: rela
tive;">⇒(3+Δ1−−−√2)2=(3−Δ1−−−√)2⇒(3+Δ12)2=(3−Δ1)2⇒3+Δ1=2(3−Δ1)⇒Δ1=1⇒Δ1=1" role="presen
tation" style="font-
size: 1
3.696px; display: inline; position: relative;">⇒3+Δ1−−−√=2(3−Δ1−−−√)⇒Δ1−−−√=1⇒Δ1=1⇒3+Δ1=2(3−Δ1)⇒Δ1=1⇒Δ1=1⇒�
394;2=2⇒Δ2=4" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇒Δ2−−−√=2⇒Δ2=4⇒Δ2=2⇒Δ2=4Ta có {Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">{Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32{Δ1=1Δ2=4⇒{9−4m=136−n=4⇔{m=2n=32 (thõa đk)