ta có \(A=x^2-5x+3=x^2-\frac{2.x.5}{2}+\frac{5^2}{4}-\frac{13}{4}=\left(x-\frac{5}{2}\right)^2-\frac{13}{4}\)
vì \(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow A\ge-\frac{13}{4}\)
dáu = xảy ra <=> x=5/2
b) ta có \(B=2x^2-4x+5=2\left(x^2-2x+\frac{5}{2}\right)\) \(=2\left(x^2-2x+1+\frac{3}{2}\right)=2\left[\left(x-1\right)^2+\frac{3}{2}\right]=2\left(x-1\right)^2+3\)
vì \(2\left(x-1\right)^2\ge0\Rightarrow B\ge3\)
dấu = xảy ra <=> x=1
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