3x²+4x-6=0
Δ'=2²-3.(-6)=22>0
⇒pt có 2 nghiệm phân biệt
Theo vi-et ta có:
$\left \{ {{x1+x2=\frac{-4}{3} } \atop {x1.x2=-2}} \right.$
a) $\frac{1}{x1}$ +$\frac{1}{x2}$
=$\frac{x1+x2}{x1.x2}$
= $\frac{-4}{3}$ : (-2)=$\frac{2}{3}$
b)$x_1^2$+$x_2^2$
=(x1+x2)²-2x1.x2
=($\frac{-4}{3}$)² -2.(-2)
=$\frac{16}{9}$ +4
=$\frac{52}{9}$
c) 3x1-x1.x2+3x2
=3(x1+x2)-x1.x2
=3.($\frac{-4}{3}$) - (-2)
=-4+2=-2
