1. Đặt: $f(x)=ax^2+bx+c$
Ta có: $f(0)=c;f\left(\dfrac{2}{3}\right)=\dfrac{4a}{9}+\dfrac{2b}{3}+c$
Khi đó: $0=\dfrac{a}{3}+\dfrac{b}{2}+c=\dfrac{3}{4}f\left(\dfrac{2}{3}\right)+\dfrac{1}{4}f(0)$
Nếu $f\left(\dfrac{2}{3}\right)=0$, suy ra đpcm.
Nếu $f\left(\dfrac{2}{3}\right)\ne0 \Rightarrow f\left(\dfrac{2}{3}\right)f(0)<0$, suy ra $f(x)=0$ có nghiệm thuộc $\left(0;\dfrac{2}{3}\right)$, đpcm.