Question

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

Answer 1

\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)

\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)

\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)

\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)

Answer 2

mình ra là \(\dfrac{11}{36}\)mà bn

bn coi lại đi

I=3x²-5x+3

I=3(x²-\(\dfrac{5}{3}\)x+1)

I=3[x²-2.x.\(\dfrac{5}{3}\)+\(\left(\dfrac{5}{6}\right)^2\)-\(\left(\dfrac{5}{6}\right)^2\)+1]

I=3(x-\(\dfrac{5}{3}\))²+\(\dfrac{11}{36}\)

I=3(x-\(\dfrac{5}{3}\))²+\(\dfrac{11}{36}\)≥\(\dfrac{11}{36}\)

vậy Min I= \(\dfrac{11}{36}\)khi x =\(\dfrac{5}{3}\)

Theo mik nghĩ là vậy á

CHÚC BN HỌC TỐT

Answer 3

Đây chỉ đơn giản là việc tách ghép để tạo hằng đẳng thức, khi đó dôi ra con số \(\frac{11}{12}\)

\(3x^2-5x+3=3(x^2-\frac{5}{3}x+1)=3(x^2-2.\frac{5}{6}x+\frac{5^2}{6^2}+\frac{11}{36})\)

\(=3(x^2-2.\frac{5}{6}x+\frac{5^2}{6^2})+3.\frac{11}{36}\)

Trong đó \(3.\frac{11}{36}=\frac{11}{12}\) đó