$x²-2(m+1)x+m²-3=0$
$(a=1;b'=-(m+1);c=m²-3)$
$Δ'=b'²-ac$
$Δ'=[-(m+1)]²-1.(m²-3)$
$Δ'=(m+1)²-m²+3$
$Δ'=m²+2m+1-m²+3$
$Δ'=2m+4$
Để $(1)$ có 2 nghiệm $x_1,x_2$
$⇔\left \{ {{a \neq 0} \atop {Δ' \geq 0}} \right.$
$⇔\left \{ {{1\neq 0 (t/m)} \atop {2m+4 \geq 0}} \right.$
$⇔2m≥-4$
$⇔m≥-2$
Vậy vs $m≥-2$ để $(1)$ có 2 nghiệm $x_1,x_2$
Theo Vi-ét ta có
$\left \{ {{x_1+x_2=\frac{-b}{a}=2m+2} \atop {x_1.x_2=\frac{c}{a}=m^2-3}} \right.$
Ta có $P=x_1(x_2+6)-3(x_1-x_2)$
$=x_1.x_2+6x_1-3x_1+3x_2$
$=m²-3+3x_1+3x_2$
$=m²-3+3(x_1+x_2)$
$=m²-3+3.(2m+2)$
$=m²-3+6m+6$
$=m²+6m+3$
$=m²+6m+9-6$
$=(m+3)²-6$ (nhận)
Vì $(m+3)²≥0;-6>-2$
Vậy.....
