\(C=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu '=' xảy ra khi x=1/2
\(D=25x^2-10xy+y^2+2y^2+4y+2-1\)
\(=\left(5x-y\right)^2+2\left(y+1\right)^2-1>=-1\)
Dấu '=' xảy ra khi y=-1 và x=1/5