\(A=x-x^2=-x^2+x=-\left(x^2-x\right)=-\left(x^2-x+1-1\right)\)
\(=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}-1\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}-1\right]=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)
Dấu "=" xảy ra <=> \(\left(x-\frac{1}{2}\right)^2=0< =>x=\frac{1}{2}\)
Vậy MaxA=1/4 khi x=1/2
\(B=-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-2.x.3+9+2\right)=-\left[\left(x-3\right)^2+2\right]=-2-\left(x-3\right)^2\le-2\)
Dấu "=" xảy ra <=> x-3=0<=>x=3
Vậy maxB=-2 khi x=3
quên,còn câu c)
\(C=4x-x^2+1=1-x^2+4x=1-\left(x^2-4x\right)=1-\left(x^2-4x+4-4\right)\)
\(=1-\left(x^2-2.x.2+4-4\right)=1-\left[\left(x-2\right)^2-4\right]=1-\left(x-2\right)^2+4=5-\left(x-2\right)^2\le5\)
Dấu "=" xảy ra \(< =>x-2=0< =>x=2\)
Vậy maxC=5 khi x=2
