A= \(x^2-x+2=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+2=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\) với mọi x
=. Min A=\(\frac{7}{4}\)khi \(\left(x-\frac{1}{2}\right)^2=0\)<=> \(x=\frac{1}{2}\)
b) B= \(3x^2-6x=\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+\left(\sqrt{3}\right)^2\right]-3\)
=> B= \(\left(\sqrt{3}x-\sqrt{3}\right)^2-3\ge-3\)
=> Min B=-3 <=> x=1
c) C= \(2x^2+4x=\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\sqrt{2}+\left(\sqrt{2}\right)^2\right]-2\)
=> C= \(\left(\sqrt{2}x-\sqrt{2}\right)^2-2\ge-2\)
=> min C=-2 khi x=1
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