Question

Cho biểu thức:

N=(\(\dfrac{x-2\sqrt{x}+1}{x-1}-\dfrac{x^2-x\sqrt{x}+x}{x\sqrt{x}+1}\)).(\(\sqrt{x}+1\)) với x≥0 và x≠1
a. Rút gọn N
b. Tìm x để N có giá trị lớn nhất

Answer 1

a) \(N=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\left(\sqrt{x}+1\right)=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{x}{\sqrt{x}+1}\right).\left(\sqrt{x}+1\right)=\dfrac{-x+\sqrt{x}-1}{\sqrt{x}+1}\left(\sqrt{x}+1\right)=-x+\sqrt{x}-1\)

b) \(N=-x+\sqrt{x}-1=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\)

\(maxN=-\dfrac{3}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)