tìm GTLN a) B= x – x^2 +6 b) C= -4y^2 + 8y -12

2 bình luận về “tìm GTLN a) B= x – x^2 +6 b) C= -4y^2 + 8y -12”

1. B = x – x^2+6

   B = – ( x^2 – x – 6 )

   B = – ( x^2 – x ) – 6

   B = – ( x^2 – 2 . x . 1/2 + 1/4 – 1/4 ) – 6

   B = – ( x – 1/2)^2 + 1/4 – 6

   B = – ( x – 1 / 2 )^2 – 23/4

   Do – ( x – 1 / 2 )^2 ≥ 0

   => – ( x – 1 / 2 )^2 – 23/4 ≥ – 23/4

   Dấu = xảy ra khi :

   ( x – 1/2 ) = 0

   x = 1/2

   Vậy max B = – 23/4 khi x = 1/2

   ————————-

   C = – 4y^2 + 8y – 12

   C = – ( 4y^2 – 8y + 12 )

   C = – ( 4y^2 – 8y ) + 12

   C = – [(2y)^2 – 2 . 2y . 2 + 4 – 4 ] + 12

   C = – ( 2y – 2 )^2 + 4 + 12

   C = – ( 2y – 2 ) ^2 + 16

   Do – ( 2y – 2 ) ^2 ≥ )
   => – ( 2y – 2 ) ^2 + 16 ≥ 16

   Dấu = xảy ra khi :

   ( 2y – 2 ) = 0

   =>2y = 2

   => y = 1

   Vậy max C = 16 khi y = 1

   Trả lời
2. a) B = x-$x^{2}$+6

   => -4B = $4x^{2}-2.2x.1+1^{2}$-25

   => -4B = $(2x-1)^{2}$-25

   Vì $(2x-2)^{2}$≥0 ∀ x

   => -4B ≥ -25 ∀ x

   => B ≤ $\frac{25}{4}$ (1)

   (1) xảy ra <=> 2x-1 = 0 <=> 2x = 1 <=> x = $\frac{1}{2}$ 

   b) C = $-4y^{2}+8y-12$ 

   => -C = $(2y)^{2}-2.2y.2+2^{2}+8$

   => -C = $(2y-1)^{2}+8$

   Vì $(2y-1)^{2}$ ≥ 0 ∀ y

   => -C ≥ 8 ∀ y

   => C ≤ -8 (2)

   (2) xảy ra <=> 2y-1=0 <=> 2y = 1 <=> y = $\frac{1}{2}$ 

   Trả lời