a) thay x=3 vào pt:
3²-2.3+m+3=0
⇔9-6+m+3=0
⇔ m= -6
Thay m= -6 vào pt
x²-2x-6+3=0
⇔x²-2x-3=0
⇔x²-3x+x-3=0
⇔x(x-3)+(x-3)=0
⇔(x-3)(x+1)=0
⇔\(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
b) Theo vi-et ta có:
$\left \{ {{x_1+x_2=2} \atop {x_1.x_2=m+3}} \right.$ (1)
Lại có: $x_1^2$+$x_2^2$-$x_1.x_2$-4=0
⇔$(x_1+x_2)^2$-2$x_1.x_2$-$x_1.x_2$-4=0
⇔$(x_1+x_2)^2$-3$x_1.x_2$-4=0 (2)
Thay (1) vào (2) ta có
2²-3.(m+3)-4=0
⇔4-3m-9-4=0
⇔ -3m-9=0
⇔ -3m=9
⇔m= -3