## tìm min A=2x^2 + 3x + 15 tìm max D= 4+2x-x^2 tìm max E= 1+3x-2x^2

Question

tìm min A=2x^2 + 3x + 15
tìm max D= 4+2x-x^2
tìm max E= 1+3x-2x^2

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3 tuần 2021-11-21T07:14:23+00:00 2 Answers 3 views 0

1. Đáp án:

A=2x^2 + 3x + 15

⇒2(x²+3/2x)+15

⇒2(x+3/4)²+111/8≥111/8

Dấu ”=” xảy ra khi :

x+3/4=0

⇒x=-3/4

⇒A_{min}=111/8 khi x= -3/4

D= 4+2x-x^2

⇒-(x²-2x-4)

⇒-(x²-2x-1)-3

⇒-(x-1)²-3≤-3

DẤu ”=” xảy ra khi :

x-1=0

⇒x=1

Vậy D_{max}=-3 khi x=1

E= 1+3x-2x^2

⇒-2(x²-3/2x)+1

⇒-2(x-3/4)²+17/8≤17/8

Dấu ”=” xảy ra khi :

x-3/4=0

⇒x=3/4

Vậy E_{max}=17/8 khi x= 3/4

2.  A = 2x^2 + 3x + 15

 = 2(x^2 + 3/(2)x) + 15

 = 2(x^2 + 2.3/(4)x + 9/16) – 9/8 + 15

 = 2.(x+3/4)^2 + 111/8 \geq 111/8

 => A_{min} = 111/8 ; khi  x = -3/4

 D  = 4 + 2x – x^2 = -(x^2 – 2x -4) = -(x^2-2x +1) -3

 = -(x-1)^2 -3 \leq -3

 => D_{max}= -3 khi  x = 1

 E =  1 + 3x – 2x^2 = -2(x^2 – 3/(2)x ) + 1

 = -2.(x^2 -3/(2)x + 9/16) + 9/8 + 1

 = -2(x-3/4)^2 + 17/8 \leq 17/8

=> E_{max} = 17/8 khi  x  = 3/4