`C=3-x^2+5x`
`=> C= -(x^2-5x-3)`
`=> C= -(x^2-5x+25/4-25/4-3)`
`=> C= -(x-5/2)^2+37/4`
Vì \(-\left(x-\dfrac{5}{2}\right)^2\le0\)
\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)
\(\Rightarrow C_{max}=\dfrac{37}{4}\)
Dấu `''=''` xảy ra khi \(-\left(x-\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow x=\dfrac{5}{2}\)