Để PT có 2 nghiệm phân biệt x1,x2x1,x2 thì:

Δ′=(m−1)2+2m>0⇔m2+1>0⇔m∈RΔ′=(m−1)2+2m>0⇔m2+1>0⇔m∈R

Áp dụng định lý Viet: {x1+x2=2(m−1)x1x2=−2m{x1+x2=2(m−1)x1x2=−2m

Khi đó:

x21+x1−x2=5−2mx12+x1−x2=5−2m

⇔x21+2x1=5−2m+x1+x2=5−2m+2(m−1)=3⇔x12+2x1=5−2m+x1+x2=5−2m+2(m−1)=3

⇔(x1+1)2=4⇔(x1+1)2=4

⇔x1+2=±2⇔x1+2=±2 ⇒[x1=0x1=−4⇒[x2=2m−2x2=2m+2⇒[x1=0x1=−4⇒[x2=2m−2x2=2m+2

⇒[x1x2=0(2m−2)x1x2=−4(2m+2)⇔[−2m=0−2m=−4(2m+2)⇒[x1x2=0(2m−2)x1x2=−4(2m+2)⇔[−2m=0−2m=−4(2m+2)

⇔[m=0m=−43