\(A=4x-x^2+3=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\)
Dấu " = " khi \(-\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(MIN_A=7\) khi x = 2
Đặt \(A=4x-x^2+3\)
\(\Leftrightarrow A=-x^2+4x+3\)
\(\Rightarrow A=-\left(x^2-4x-3\right)\)
\(\Rightarrow A=-\left(x^2-4x+4-7\right)\)
\(\Rightarrow A=\left(x-2\right)^2+7\le7\)
Vậy Amax=7
Ta có: \(4x-x^2+3=-\left(x^2-4x\right)+3=-\left(x^2+2.x.2+4\right)+4+3\)
\(=-\left(x^2+2\right)+7\)
Do \(-\left(x^2+2\right)^2\le0\) với mọi x \(\Rightarrow-\left(x+2\right)^2+7\le7\) hay \(4x-x^2+3\le7\)
Dấu ''='' xảy ra khi \(-\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)
Vậy giá trị lớn nhất của \(4x-x^2+3\) là 7 khi x=-2
\(A=4x-x^2+3=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-1\right)\)
\(=-\left(x+2\right)^2+1\le1\)
Dấu " = " khi \(-\left(x+2\right)^2=0\Rightarrow x=-2\)
Vậy \(MIN_A=1\) khi x = -2
