Question

[Hồ Trang]

1:Tìm GTNN của biểu thức

D=-5/x^2-4x+7

2;Tìm GTLN của biểu thức

E=2x^2+4x+4/x^2+2x+4

F=6x+8/x^2+1

Giúp mik vs ạ!Mik cần gấp

Answer 1

[Kien Nguyen]

D = \(-\dfrac{5}{x^2-4x+7}\)

Vì: x² - 4x + 7

= x² - 4x + 4 + 3

= (x - 2)² + 3 \(\ge\) 3 \(\forall\)x

\(\Rightarrow\) \(\dfrac{5}{\left(x-2\right)^2+3}\) \(\le\) \(\dfrac{5}{3}\) \(\forall\)x

\(\Rightarrow\) \(-\dfrac{5}{\left(x-2\right)^2+3}\)\(\ge\)-\(\dfrac{5}{3}\) \(\forall\)x

Dấu"=" xảy ra khi:

x - 2 = 0

\(\Rightarrow\) x = 2

Vậy.............

E = \(\dfrac{2x^2+4x+4}{x^2+2x+4}\)

Ta có:

\(\dfrac{2x^2+4x+4}{x^2+2x+4}\)

= \(\dfrac{2\left(x^2+2x+4\right)-4}{x^2+2x+4}\)

= 2 - \(\dfrac{4}{x^2+2x+4}\)

Vì:

x² + 2x + 4

= x² + 2x + 1 + 3

= (x + 1)² + 3 \(\ge\) 3 \(\forall\)x

\(\Rightarrow\) \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{4}{3}\) \(\forall\)x

\(\Rightarrow\) 2 - \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{2}{3}\) \(\forall\)x

Dấu "=" xảy ra khi:

x + 1 = 0

\(\Rightarrow\) x = -1

Vậy...............

F = \(\dfrac{6x+8}{x^2+1}\)

= \(\dfrac{x^2+6x+9-x^2-1}{x^2+1}\)

= \(\dfrac{\left(x+3\right)^2-\left(x^2+1\right)}{x^2+1}\)

= \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\) \(\ge\) -1 \(\forall\)x

Dấu "=" xảy ra khi:

(x + 3)² = 0

\(\Rightarrow\) x + 3 = 0

\(\Rightarrow\) x = -3

Vậy.....................

Answer 2

[Hồ Trang]

Cảm ơn bạn nha🙂