\(D=x+\sqrt{x^2+\frac{1}{x^2}+2}\)    Với x < 0

 

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22 tháng 9 2020

đk: \(x\ne0\)

Ta có:

\(D=x+\sqrt{x^2+\frac{1}{x^2}+2}\)

\(D=x+\sqrt{\left(x+\frac{1}{x}\right)^2}\)

\(D=x+\left|x+\frac{1}{x}\right|\)

\(D=x-x-\frac{1}{x}\) \(\left(x< 0\right)\)

\(D=-\frac{1}{x}\)

31 tháng 7 2019

#)Giải :

Bài 1 :

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

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\frac{\left(1-x\right)^2}{2}\)

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

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

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

b) Để \(P>0\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)

c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu ''='' xảy ra khi \(x=\frac{1}{4}\)

27 tháng 8 2016

Ta có: 

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

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

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

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

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

\(P=\sqrt{x}-x\)

b) Để \(P>0\) thì \(\sqrt{x}-x>0\)

  • \(\sqrt{x}-x>0\)

   \(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Suy ra: TH1\(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)

            TH2:\(\sqrt{x}>0\)  và \(1-\sqrt{x}>0\) (Nhận)

Ta có \(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) để \(P>0\)

  • \(\sqrt{x}>0\) \(\Rightarrow x>0\)
  • \(1-\sqrt{x}>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)

Vậy để \(P>0\) thì \(0< x< 1\)

c)\(P=\sqrt{x}-x\)

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

\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)

\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)

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

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)

Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)

Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)