Đáp án+Giải thích các bước giải:
$a)M=4x-x^2+3\\\ \ \ \ \ \ \ \ = -x^2+2.2x-4+7\\\ \ \ \ \ \ \ \ = -(x^2-2.2x+2^2)+7\\\ \ \ \ \ \ \ \ = -(x+2)^2+7\\Vì \ -(x+2)^2\leq0\\$ `-(x-2)^2+7<=7`
Vậy `Max_{M}=7` tại
`x-2=0=>x=2`
`b) N=x-x^2`$\\$`\ \ \ \ \ \ \ =-x^2+x-(1/2)^2+1/4`$\\$`\ \ \ \ \ \ \ =-[x^2-x+(1/2)^2]+1/4`$\\$`\ \ \ \ \ \ \ =-(x-1/2)^2+1/4`$\\$`Vì\ -(x-1/2)^2 <=0`$\\$`=>-(x-1/2)^2+1/4<1/4`$\\$`Vậy\ Max_{N}=1/4\ tại`$\\$`x-1/2=0=>x=1/2`
`c)P=2x-2x^2-5`$\\$`\ \ \ \ \ \ \ =-2x^2+2x-1/2-9/2`$\\$`\ \ \ \ \ \ \ =-2[x^2-x+(1/2)^2]-9/2`$\\$`-2.(x-1/2)^2-9/2`$\\$`Vì\ -2(x-1/2)^2 <=0`$\\$`=>-2(x-1/2)^2-9/2<-9/2`$\\$`Vậy\ Max_{P}=-9/2\ tại`$\\$`x-1/2=0=>x=1/2`
