Question

P= \((\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}). \dfrac{(1-x)^2}{2}\)

a) tìm Tập xác định, rút gọn P

b) c/m nếu 0<x<1=> P>0

c) Tìm P max

 

 

Answer 1

a: ĐKXĐ: x>=0; x<>1

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

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

\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{2}\cdot\dfrac{x-1}{\sqrt{x}+1}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b: 0<x<1

=>căn x<1

=>căn x-1<0

=>căn x*(căn x-1)<0

=>-căn x*(căn x-1)>0

=>P>0

c: \(P=-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}\)

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

Dấu = xảy ra khi x=1/4