Định m để phương trình sau có bốn nghiệm phân biệt:
x2+x=m|2x2|
x2+x=m|2x2|
[x2+x=2mx2m,x1x2+x=2mx+2m,x<1[x2(2m1)x+2m=0,x1x2(2m1)x+2m=0,x<1
+/ phương trình x2(2m1)x+2m=0Δ=4m24m+18m=4m212m+1

Với Δ>0m<322 hay m>32+2, phương trình có hai nghiệm

x1=(2m1)4m212m+12,x2=(2m1)+4m212m+12
 Cần (2m1)4m212m+121(2m1)4m212m+12
2m34m212m+1{x>324m212m+94m212m+1m32
phối hợp với Δ>0(1)
+/ phương trình x2+(2m1)x2m=0Δ=4m2+4m+1+8m=4m2+12m+1

Với  Δ>0m<322 hay m>32+2, phương trình có hai nghiệm

x3=(2m+1)4m2+12m+12,x4=(2m+1)+4m2+12m+12

  Cần (2m+1)+4m2+12m+121(2m1)+4m212m+12
2m+34m2+12m+1
{x>324m2+12m+14m2+12m+9m>m>32
phối hợp với Δ>0(2)
(2m1)±4m212m+12(2m+1)±4m2+12m+12
 4m±4m212m+14m2+12m+1
 16m2+4m212m+1±8m4m212m+14m2+12m+1
 16m224m8m4m212m+1
 2m23m±4m212m+1
4m4+9m212m39m412m3+m2
 m0(3)
 Hợp (1),(2), giao với (3) được m>32+2m0

Thẻ

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