Cho phương trình x^2+4x+m+1=0 (với m là tham số) a) Giải phương trình (1) với m = 2

x2+4x+m+1=0  (1)

a) khi m=2 (1) thành  x2+4x+3=0⇔x2+3x+x+3=0

⇔x(x+3)+(x+3)=x+1x+3⇔x=−1x=−3

b) Ta có: Δ'=(−2)2−m−1=3−m

c) vì m<3, ¸áp dụngVi et⇒x1+x2=−4x1x2=m+1⇔x2=−4−x1

Tac ó: x1−12x2−x2−12x1=−3

⇔x1−12(−4−x1)−−4−x1−12x1=−3

⇔x1(x1−1)−(−5−x1)(−4−x1)2x1(−4−x1)=−3

⇔x12−x1−(5+x1)(4+x1)=−3.2x1(−4−x1)

⇔x12−x1−20−9x1−x12=24x1+6x12⇔6x12+34x1+20=0⇔x=−23x=−5

*) x1=−23⇒x2=−103⇒m+1=209⇔x=119*) x2=−5⇒x2=1⇒m+1=−5⇔m=−6

Vậy m=−6; m=119