Question

Tìm `min:`

`C=9x^2+5-6x`

`D=1+x^2-x`

Answer 1

\(C=\left(9x^2-6x+1\right)+4=\left(3x-1\right)^2+4\ge4\)

\(C_{min}=4\) khi \(x=\dfrac{1}{3}\)

\(D=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(D_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

Answer 2

\(C=9x^2+5-6x=\left(9x^2-6x+1\right)+4=\left(3x-1\right)^2+4\ge4\)

\(minC=4\Leftrightarrow x=\dfrac{1}{3}\)

\(D=1+x^2-x=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minD=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)