Question

[Nguyễn Viết Tùng]

 

Answer 1

[GV Nguyễn Trần Thành Đạt]

Bài 3:

\(a,A=x^2-x=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\in R\\ Vậy:min_A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(b,B=2x^2-8x=2\left(x^2-x\right)=2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{1}{2}=2.\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\forall x\in R\\ Vậy:min_B=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(c,C=3x^2-12x=3\left(x^2-x\right)=3\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ Vậy:min_C=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)