Question

tìm GTLN hoặc GTNN:

a)A= 2x² - 8x - 10x b)B= 9x - 3x²

c) C= x² - 6x + 11 d) D= 2x² + 10x - 1

e) E= 5x - x²

GIÚP MK VS

Answer 1

D= 2( \(x^2\)+5x-\(\dfrac{1}{2}\))

D= 2( \(x^2\)+ 2. \(\dfrac{5}{2}\)x + \(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))

D= 2( x+\(\dfrac{5}{2}\))\(^2\)+ \(\dfrac{27}{8}\) lớn hơn hoặc bằng \(\dfrac{27}{8}\)

vậy min P = \(\dfrac{27}{8}\) <=> x = -\(\dfrac{5}{2}\)

Answer 2

e)\(E=5x-x^2=-x^2+5x=-x^2+2\cdot x\cdot\dfrac{5}{2}-\dfrac{25}{4}+\dfrac{25}{4}=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

(Vì: \(\left(x-\dfrac{5}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{5}{2}\right)^2\le0\))

Vậy \(MaxE=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

Answer 3

b)\(B=9x-3x^2=-3\left(x^2-3x\right)=-3\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{27}{4}=-3\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\le\dfrac{27}{4}\)

( Vì : \(\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow-3\left(x-\dfrac{3}{2}\right)^2\le0\))

Vậy \(MaxB=\dfrac{27}{4}\) khi \(x=\dfrac{3}{2}\)

Answer 4

d) \(D=2x^x+10x-1=2\left(x^2+5-\dfrac{1}{2}\right)=2\left(x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{27}{4}\right)=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\)

(Vì: \(\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow2\left(x+\dfrac{5}{2}\right)^2\ge0\))

Vậy \(MinD=-\dfrac{27}{2}\) khi \(x=-\dfrac{5}{2}\)

Answer 5

c) C= x\(^2\)- 2.3x + 9+3

C= (x-3)\(^2\)+3 lớn hơn hoặc bằng 3

=> Min C = 3 <=> x= 3

Answer 6

a)Sửa đề: \(2x^2-8x-10\)

\(=2x^2-8x+8-18\)

\(=2\left(x^2-4x+4\right)-18\)

\(=2\left(x-2\right)^2-18\ge-18\) (Vì: \(\left(x-2\right)^2\ge0\Rightarrow2\left(x-2\right)^2\ge0\))

Vậy GTNN của \(2x^2-8x-10\) là \(-18\)

Answer 7

c)\(C=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\) (Vì:\(\left(x-3\right)^2\ge0\))

Vậy \(MinC=2\) khi \(x=3\)