Question

Tìm giá trị lớn nhất

M=2x^2+3x+5

Answer 1

=2(x^2+3/2x+5/2)

=2(x^2+2*x*3/4+9/16+31/16)

=2(x+3/4)^2+31/8>=31/8

Dấu = xảy ra khi x=-3/4

Answer 2

`M=2x^2+3x+5`

`M=2(x^2+3/2x+5/2)`

`M=2(x^2+2. x. 3/4+9/16+31/16)`

`M=2(x+3/4)^2+31/8`

Vì `2(x+3/4)^2 >= 0 AA x<=>2(x+3/4)^2+31/8 >= 31/8 AA x`

     Hay `M >= 31/8 AA x`

Dấu "`=`" xảy ra `<=>x+3/4=0<=>x=-3/4`