a/ C1: Do ac=2.(-2)<0 => pt luôn có 2 ng phân biệt
C2: \(\Delta=\left(-3m\right)^2-4.2.\left(-2\right)\)
\(=9m^2+16\ge16\)
=> pt luôn có 2 ng phân biệt
b/ Có \(\hept{\begin{cases}x_1+x_2=\frac{3m}{2}\\x_1.x_2=-1\end{cases}}\) (vi-et)
\(\Rightarrow x+x=\left(x+x\right)^2-2xx\)
\(=\left(\frac{3m}{2}\right)^2-2.\left(-1\right)\)
\(=\frac{9m^2}{4}+2\ge2\)
Vậy min=2 <=> m=0
c\(\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x^3_1+x^3_2}{x^3_1x^3_2}\)
= \(\frac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{x_1^3x_2^3}\)
\(=\frac{\left(\frac{3m}{2}\right)^2-3\left(-1\right)\left(\frac{3m}{2}\right)}{\left(-1\right)^3}\)
\(=\frac{\frac{9m^2}{4}+\frac{9m}{2}}{-1}\)
\(=\frac{\frac{9m^2}{4}+\frac{18m}{4}}{-1}\)
\(=\frac{9m^2+18m}{-4}\)
