a, Ta có : \(x=81\Rightarrow\sqrt{x}=9\)

Thay \(\sqrt{x}=9\)vào biểu thức A ta được : 

\(A=\frac{2}{9+1}=\frac{2}{10}=\frac{1}{5}\)

b, Ta có : \(P=\frac{B}{A}\)hay\(P=\frac{\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}}{\frac{2}{\sqrt{x}+1}}\)

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

c, Ta có \(\frac{1}{2}=\frac{\sqrt{x}}{2\sqrt{x}}\)mà \(\sqrt{x}< \sqrt{x}+1\)

nên \(P>\frac{1}{2}\)

a) \(A=\frac{2}{\sqrt{x}+1}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}\)

b) \(B=\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}\)

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

\(\Rightarrow P=\frac{B}{A}=\frac{1}{\sqrt{x}}\div\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c) Ta có: \(P=\frac{\sqrt{x}+1}{2\sqrt{x}}=\frac{1}{2}+\frac{1}{\sqrt{x}}+\frac{1}{2}+0=\frac{1}{2}\)

=> P>1/2

a)Thay $x=81(\mathrm{t}\mathrm{m}dk)$ vào biều thức $A$, ta được \mathrm{A}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}A=81​+12​=9+12​=51​
Vậy $x=81$ thì \mathrm{A}=\frac{1}{5}A=51​

b) B=\frac{1}{\sqrt{x}(\sqrt{x}+1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)1​+x​(x​+1)x​​
B=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)x​+1​
B=\frac{1}{\sqrt{x}}B=x​1​
P=\frac{1}{\sqrt{x}} \div \frac{2}{\sqrt{x}+1}P=x​1​÷x​+12​
P=\frac{1}{\sqrt{x}} \cdot \frac{\sqrt{x}+1}{2}P=x​1​⋅2x​+1​
P=\frac{\sqrt{x}+1}{2 \sqrt{x}}P=2x​x​+1​

c) Ta có P-\frac{1}{2}=\frac{\sqrt{x}+1}{2 \sqrt{x}}-\frac{1}{2}=\ldots=\frac{1}{2 \sqrt{x}}P−21​=2x​x​+1​−21​=…=2x​1​
Ta có $x>0$ nên \frac{1}{2 \sqrt{x}}>0 \Rightarrow P-\frac{1}{2}>0 \Rightarrow P>\frac{1}{2}2x​1​>0⇒P−21​>0⇒P>21​
 

\(ĐKXĐ:x\ne1;x\ne0\)

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

\(N=\frac{\sqrt{x}-3}{2\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)2\sqrt{x}}=\frac{x-2\sqrt{x}-3}{2x+2\sqrt{x}}\)

Ta có :

 \(x\ge0>-3\)

\(\Leftrightarrow x>-3\)

\(\Leftrightarrow x+\left(x-2\sqrt{x}\right)>-3+\left(x-2\sqrt{x}\right)\)

\(\Leftrightarrow2x-2\sqrt{x}>x-2\sqrt{x}-3\)

\(\Leftrightarrow\frac{2x-2\sqrt{x}}{2x+2\sqrt{x}}>\frac{x-2\sqrt{x}-3}{2x+2\sqrt{x}}\)

\(\Leftrightarrow A>N\)

Ta có: \(P=\frac{\sqrt{x}-4}{\sqrt{x}}\times\frac{x+\sqrt{x}+1}{\sqrt{x}-4}\)

 \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)\(\left(ĐK:x>0\right)\)

Ta lấy \(P-2=\frac{x+\sqrt{x}+1}{\sqrt{x}}-2\)

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

                       \(=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

                       \(=\frac{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{3}{4}}{\sqrt{x}}\)

                      \(=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}\)

Vì \(x>0\Rightarrow\sqrt{x}>0\)

 \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}>0\)

\(\Rightarrow P-2>0\)

\(\Rightarrow P>2\)

Học tốt 

Mau la \(\sqrt{X - 3} \) that sao

????

1) Thay x=16 vào A ta có:

A=\(\frac{16+\sqrt{16}+1}{\sqrt{16}+2}\)

A=\(\frac{16+4+1}{4+2}\)

A=\(\frac{21}{6}=\frac{7}{2}\)

\(2,\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x-\sqrt{x}+2}{x-\sqrt{x}}\)

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

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

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

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

\(3,P=A.B=\frac{x+\sqrt{x}+1}{\sqrt{x}+2}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Ta thấy \(\left(\sqrt{x}-1\right)^2>0\Rightarrow x-2\sqrt{x}+1>0\)

\(\Rightarrow x+\sqrt{x}+1>3\sqrt{x}\)

\(\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>\frac{3\sqrt{x}}{\sqrt{x}}\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}>3\left(đpcm\right)\)