\(B=9x^2-6x+5=9x^2-6x+1+4\\ B=\left(3x-1\right)^2+4\ge4\)
đẳng thức xảy ra khi 3x-1=0 => x=1/3
vậy min B=4 tại x=1/3
\(C=x^2+x-3\)
\(C=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{2}^2-\dfrac{1}{2}^2-3\)
\(C=\left(x+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\)
Ta có: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow C\ge-\dfrac{13}{4}\)
Vậy MinC=-13/4 khi x=-1/2
\(D=2x^2+2xy+y^2-2x+2y+2\)
\(D=\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+x^2-4x+1\)
\(D=\left(x+y+1\right)^2+\left(x-2\right)^2-3\)
MinD=-3 khi x=2; y=-3
B=9x2-6x+5
=9
= 9\(\left(x-\dfrac{1}{3}\right)^2+4\)
