a)\(A=4x-x^2+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x^2-4x+4\right)+7\)
\(=-\left(x-2\right)^2+7\le7\)
Dấu = khi \(x=2\)
Vậy MaxA=7 khi \(x=2\)
b)\(B=x-x^2\)
\(=-\left(x^2-x\right)\)
\(=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)\)
\(=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu = khi \(x=\frac{1}{2}\)
Vậy MaxB=\(\frac{1}{4}\)khi \(x=\frac{1}{2}\)
\(A=4x-x^2+3=7-x^2+4x-4=7-\left(x-2\right)^2\le7\)
\(MaxA=7\Leftrightarrow x=2\)
\(B=x-x^2=\frac{5}{4}-x^2+x-\frac{1}{4}=\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{5}{4}\)
\(MaxB=\frac{5}{4}\Leftrightarrow x=\frac{1}{2}\)
\(N=2x-2x^2-5=-\frac{9}{2}-2x^2+2x-\frac{1}{2}=-\frac{9}{2}-2\left(x-\frac{1}{4}\right)^2\le-\frac{9}{2}\)
\(MaxN=-\frac{9}{2}\Leftrightarrow x=\frac{1}{4}\)
