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a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
#)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 \(x=\frac{1}{4}\)
a, Ta có : \(A=\frac{\sqrt[]{x}-2}{x+\sqrt{x}+1};x=16\Rightarrow\sqrt{x}=4\)
\(A=\frac{4-2}{16+4+1}=\frac{2}{21}\)
b, Với \(x\ge0;x\ne1\)ta có :
\(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt[]{x}}\)
\(=\frac{x+2}{\left(\sqrt{x}\right)^2-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\)\(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\) \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(-\frac{\sqrt{x}+x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+x+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
= \(\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+x+1}\)
học tốt
\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(A=\frac{x+2}{\sqrt{x}^3-1^3}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có : x + 1 \(\ge\)\(2\sqrt{x}\)nên \(x+\sqrt{x}+1\ge3\sqrt{x}\)
\(\Rightarrow A=\frac{\sqrt{x}}{x+\sqrt{x}+1}\le\frac{\sqrt{x}}{3\sqrt{x}}=\frac{1}{3}\)
Vậy GTLN của A là \(\frac{1}{3}\)\(\Leftrightarrow x=1\)