Vì a + b = 0 => a = -b
Ta có f(3) = a.32 + b.3 + c
= 9a + 3b + c
= 9(-b) + 3b + c
= -6b + c
f(-2) = a.(-2)2 + b(-2) + c
= 4a - 2b + c
= 4(-b) - 2b + c
= -6b + c
Khi đó f(3).f(-2) = (-6b + c)(-6b + c) = (-6b + c)2 \(\ge\)0 (đpcm)
Xét đa thức \(f\left(x\right)=ax^2+bx+c\)
\(f\left(3\right)=9a+3b+c\)
\(f\left(-2\right)=4a-2b+c\)
\(\Rightarrow f\left(3\right)-f\left(-2\right)=9a+3b+c-\left(4a-2b+c\right)=9a+3b+c-4a+2b-c\)\(=5a+5b=5\left(a+b\right)=5.0=0\) (vì \(a+b=0\))
\(\Rightarrow f\left(3\right)=f\left(-2\right)\)
\(f\left(3\right).f\left(-2\right)=\left[f\left(3\right)\right]^2\)
Vì\(\left[f\left(3\right)\right]^2\ge0\) nên \(f\left(3\right).f\left(-2\right)\ge0\) (đpcm)
thanh kiu các bn ngeng