f: Ta có: \(F=x^2-xy+y^2-3y+6\)
\(=x^2-xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2-2\cdot\dfrac{\sqrt{3}}{2}y\cdot\sqrt{3}+3+3\)
\(=\left(x-\dfrac{1}{2}y\right)^2+\left(\dfrac{\sqrt{3}}{2}y-\sqrt{3}\right)^2+3\ge3\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(1;2\right)\)
a: Ta có: \(A=4x^2-4x-3\)
\(=4x^2-4x+1-4\)
\(=\left(2x-1\right)^2-4\ge-4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
b: Ta có: \(B=x^2-5x+1\)
\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{21}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
c: Ta có: \(C=4x^2+5x+3\)
\(=4\left(x^2+\dfrac{5}{4}x+\dfrac{3}{4}\right)\)
\(=4\left(x^2+2\cdot x\cdot\dfrac{5}{8}+\dfrac{25}{64}+\dfrac{23}{64}\right)\)
\(=4\left(x+\dfrac{5}{8}\right)^2+\dfrac{23}{16}\ge\dfrac{23}{16}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{8}\)
