Question

Bài 1 :

a) Tìm GTNN b) Tìm GTLN
A= x + \(\sqrt{x}\) M= -x + \(\sqrt{x-1}\)

B= x - \(\sqrt{x}\) N= (1 - 2\(\sqrt{x}\)) (\(\sqrt{x}\) - 2 )

C= x - \(\sqrt{x-2005}\) D= \(\dfrac{1}{x+\sqrt{x-1}}\)

D = x + 2\(\sqrt{x-2}\)

Answer 1

a: \(A=x+\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=0\)

Dấu '=' xảy ra khi x=0

b: \(B=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}< =-\dfrac{1}{4}\)

Dấu = xảy ra khi x=1/4

c: \(=x-2005-\sqrt{x-2005}+2005\)

\(=\left(\sqrt{x-2005}\right)^2-2\cdot\sqrt{x-2005}\cdot\dfrac{1}{2}+\dfrac{1}{4}+2004.75\)

\(=\left(\sqrt{x-2005}-\dfrac{1}{2}\right)^2+2004.75>=2004.75\)

Dấu '=' xảy ra khi x=2005,25

d: \(D=x-2+2\sqrt{x-2}+2\)

\(=\left(\sqrt{x-2}+1\right)^2+1>=2\)

Dấu '=' xảy ra khi x=2