Question

Tìm giá trị nhỏ nhất của : 3x^2+x-1

 

Answer 1

\(=3\left(x^2+\dfrac{1}{3}x-\dfrac{1}{3}\right)\\ =3\left(x^2+2\cdot\dfrac{1}{6}x+\dfrac{1}{36}-\dfrac{13}{36}\right)\\ =3\left(x+\dfrac{1}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{6}\)

Answer 2

A=3x^2+x-1

A=3x^2+1/3+2/3

A=3(x^2-2x/3+1/9)+2/3

A=3(x-1/3)^3+2/3_>2/3