\(M=4x-x^2+3\)
\(=-x^2+4x+3\)
\(=-x^2+4x-4+7\)
\(=-\left(x^2-4x+4\right)+7\)
\(=-\left(x-2\right)^2+7\)
Vì; \(-\left(x-2\right)^2+7\le7\forall x\)
=> Max M =7 tại \(-\left(x-2\right)^2=0\Rightarrow x=2\)
Ta có: \(N=x-x^2=-x^2+x\)
\(=-x^2+x-\frac{1}{4}+\frac{1}{4}\)
\(=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì: \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
=> Max N =1/4 tại \(-\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
=.= hok tốt!!
a)\(M=4x-x^2+3\)
\(M=-x^2+4x+3\)
\(M=-x^2+4x-4+7\)
\(M=-\left(x-2\right)^2+7\le7.Với\forall x\in Q\)
Dấu "=" xảy ra khi x = 2
Vậy Max M = 7 <=> x = 2
b)\(N=x-x^2=-x^2+x\le x\)
Dấu "=" xảy ra khi x = 0
=> Max N = 0 <=> x = 0
