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[myangel25697]

cho $ P=( \dfrac{ \sqrt{x}-2 }{x-1} - \dfrac{ \sqrt{x}+2 }{x+2 \sqrt{x}+1 } ). \dfrac{(1-x)^2}{2} $
a)rút gọn P
b)
c/m nếu 0<x<1 thì P>0
c)tìm Pmax

[angleofdarkness]

a/

 

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

 

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

 

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

 

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

 

$=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\sqrt{x}+1}. \dfrac{x-1}{2}$

 

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

 

b/

 

Theo a ta có $P=\dfrac{-2\sqrt{x}(\sqrt{x}-1)}{2}$ nên P > 0 khi $\sqrt{x}(\sqrt{x}-1)$ < 0.

 

Giải BPT trên được o < x < 1.

 

c/

 

Áp dụng BĐT Cauchy cho 2 số không âm ta tìm được max P.

 

Edited by angleofdarkness, 27-12-2013 - 18:48.