Question

tính GTNN của a) C=x^2-x

b)D=16x^2+2x

c)5x^2+4y^2-4x-4xy+3

 

Answer 1

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

\(C_{min}=-\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)

\(D=\left(16x^2+2x+\dfrac{1}{16}\right)-\dfrac{1}{16}=\left(4x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\ge-\dfrac{1}{16}\)

\(D_{min}=-\dfrac{1}{16}\) khi \(x=-\dfrac{1}{16}\)

\(E=\left(x^2-4xy+4y^2\right)+\left(4x^2-4x+1\right)+2\)

\(E=\left(x-2y\right)^2+\left(2x-1\right)^2+2\ge2\)

\(E_{min}=2\) khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)