a: =x^2-2x+1+4
=(x-1)^2+4>=4
Dấu = xảy ra khi x=1
b: B=x^2+4x+4+3
=(x+2)^2+3>=3
Dấu = xảy ra khi x=-2
a) \(A=x^2-2x+5\)
\(A=x^2-2x+1+4\)
\(A=\left(x-1\right)^2+4\)
Mà: \(\left(x-1\right)^2\ge0\) nên \(C=\left(x-1\right)^2+4\ge4\)
Dấu " = " xảy ra:
\(\left(x-1\right)^2+4=4\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy: \(A_{min}=4\) khi \(x=1\)
b) \(B=x^2+4x+7\)
\(B=x^2+4x+4+3\)
\(B=\left(x+2\right)^2+3\)
Mà: \(\left(x+2\right)^2\ge0\) nên \(B=\left(x+2\right)^2+3\ge3\)
Dấu " = " xảy ra:
\(\left(x+2\right)^2+3=3\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy: \(B_{min}=3\) khi \(x=-2\)
