Đáp án:
PT:$x^2+mx+1=0$
$CóΔ=m^2-4$
Để pt có 2 nghiệm $x_{1};x_{2}$ thì
$Δ≥0$
$⇔m^2-4≥0$
$⇔\left[ \begin{array}{l}m≤-2\\m≥2\end{array} \right.$
Theo hệ thức Viets ta có:
$\left \{ {{x_{1}+x_{2}=-m} \\ {x_{1}.x_{2}=1}} \right.$
$a)x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}.x_{2}=m^2-2$
$b)x_{1}^3+x_{2}^3=(x_{1}+x_{2})^3-3x_{1}.x_{2}(x_{1}+x_{2})=-m^3+3m$
$c)P=|x_{1}-x_{2}|$
$⇒P^2=x_{1}^2-2x_{1}.x_{2}+x_{2}^2=(x_{1}+x_{2})^2-4x_{1}.x_{2}=m^2-4$
$⇒P=\sqrt{m^2-4}$
