Question

À = 4x-x^2+3 tìm GTLN

Answer 1

\(A=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(A_{max}=7\) khi \(x=2\)

Answer 2

Ta có:A=4x-x²+3=-(x²-4x+4)+7=-(x-2)²+7

Vì -(x-2)²≤0

=> A≤7

Dấu "=" xảy ra <=> x=2

Answer 3

Đặt  \(A=4x-x^2+3\)

\(=-x^2+4x+3=-\left(x^2-4x-3\right)\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left[\left(x-2\right)^2-7\right]\)

\(=-\left(x-2\right)^2+7\)

Ta có:  \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2+7\le7\)

Dấu "=" khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

 vậy \(MAX_A=7\)  khi x=2