\(a=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
\(b=x^2-6x-10=x^2-6x+9-19=\left(x-3\right)^2-19\ge-19\)
\(c=16x^2+16x+7=16x^2+16x+4+3=\left(4x+2\right)^2+3\ge3\)
a)A=\(x^2-2x+5\)
\(=x^2-2x.1+1+4\)
\(=\left(x^2-2x.1+1\right)+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\in R\)
Vì \(\left(x-1\right)^2\ge0\forall x\in R\)
Dấu ''='' xảy ra khi x-1=0\(\Rightarrow x=1\)
Vậy A\(_{min}\)=4 khi x=1
b) B\(=x^2-6x-10\)
\(=x^2-2x.3+3^2-3^2-10\)
\(=\left(x-3\right)^2-19\ge-19\forall x\in R\)
Vì \(\left(x-3\right)^2\ge0\forall x\in R\)
Dấu ''='' xảy ra khi x-3=0 => x=3
Vậy \(B_{min}\)\(=-19\) khi x=3
c)C\(=16x^2+16x+7\)
\(=16\left(x^2+x+\dfrac{7}{16}\right)\)
\(=16\left(x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{16}\right)\)
\(=16\left[\left(x+\dfrac{1}{2}\right)+\dfrac{3}{16}\right]\)
\(=16\left(x+\dfrac{1}{2}\right)+3\ge3\forall x\in R\)
Vì \(16\left(x+\dfrac{1}{2}\right)\ge0\forall x\in R\)
Dấu''='' xảy ra khi x+\(\dfrac{1}{2}=0\) => x=\(\dfrac{-1}{2}\)
Vậy \(C_{min}\)=3 khi x=\(\dfrac{-1}{2}\)
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