m(x²+x+1)=x²+x+2
(m-1)x²+(m-1)x+(m-2)=0
Δ=(m-1)²-4(m-1)(m-2)
=m²-2m+1-4m²+12m-8
=-3m²+10m-7
有2个相等实根,Δ=0
-3m²+10m-7=0
(-3m+7)(m-1)=0
-3m+7=0
m=7/3
或者m-1=0
m=1
∵当m=1时,方程变为:1=2
∴m=1舍去
因此,m=7/3
m(x²+x+1)=x²+x+2
(m-1)x²+(m-1)x+(m-2)=0
Δ=(m-1)²-4(m-1)(m-2)
=m²-2m+1-4m²+12m-8
=-3m²+10m-7
有2个相等实根,Δ=0
-3m²+10m-7=0
(-3m+7)(m-1)=0
-3m+7=0
m=7/3
或者m-1=0
m=1
∵当m=1时,方程变为:1=2
∴m=1舍去
因此,m=7/3