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

0 bình luận về “tìm min A=2x^2 + 3x + 15 tìm max D= 4+2x-x^2 tìm max E= 1+3x-2x^2”

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`

    

   Trả lời

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`

    

   Trả lời