I=3x2-5x+3
I=3(x2-\(\dfrac{5}{3}\)x+1)
I=3(x2-2x.\(\dfrac{5}{6}\)+\(\dfrac{25}{36}\))+\(\dfrac{11}{12}\)
I=3(x-\(\dfrac{5}{6}\))2+\(\dfrac{11}{12}\) \(\ge\)\(\dfrac{11}{12}\)
vậy Min I=\(\dfrac{11}{12}\) khi x =\(\dfrac{5}{6}\)
cho mình hỏi I=3(x2-2x.\(\dfrac{5}{6}\)+\(\dfrac{25}{36}\))+\(\dfrac{11}{12}\) khúc này tính làm sao mà được \(\dfrac{11}{12}\) vậy