\(A=4x-x^2+3=-\left(x^2-2\times x\times2+4-4-3\right)=-\left[\left(x-2\right)^2-7\right]\)
\(\left(x-2\right)^2\ge0\)
\(\left(x-2\right)^2-7\ge-7\)
\(-\left[\left(x-2\right)^2-7\right]\le7\)
Vậy Max A = 7 khi x = 2.
\(B=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)
Vậy Max B = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)
a) \(A=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\)
Vì: \(-\left(x-2\right)^2\le0\)
=> \(-\left(x-2\right)^2+7\le7\)
Vậy GTLN của A là 7 khi x=2
b) \(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì: \(-\left(x-\frac{1}{2}\right)^2\le0\)
=> \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Vậy GTLN của B là \(\frac{1}{4}\) khi \(x=\frac{1}{2}\)
a)A=4x-x2+3
Ta có:A=4x-x2+3
A=-(x2-2.2x+4)+7
A=7-(x-2)2
Vì -(x-2)2\(\le\)0
Suy ra:7-(x-2)2\(\le\)7
Dấu = xảy ra khi x-2=0
x=2
Vậy Max A=7 khi x=2
b)B=x-x2
Ta có:B=x-x2
B=-(x2-2.\(\frac{1}{2}\)x+\(\frac{1}{4}\))+\(\frac{1}{4}\)
\(B=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\)
Vì \(-\left(x-\frac{1}{2}\right)^2\le0\)
Suy ra:\(\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)
Dấu = xảy ra khi \(x-\frac{1}{2}=0;x=\frac{1}{2}\)
Vậy Max B=\(\frac{1}{4}\) khi x=\(\frac{1}{2}\)
