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

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

2, Với x>1 ta có \(\frac{1}{A}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}\)

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

Áp dụng bđt AM-GM ta có

\(\frac{1}{A}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{3}{\sqrt{x}-1}}+3=2\sqrt{3}+3\)

Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)

\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)

\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

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

         \(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)

        \(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)

       \(=-3\)

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

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

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

    \(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, Ta có \(B< A\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)

\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)

\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)

Vậy ...

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

ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

Để A > 0 

=> \(\frac{\sqrt{x}-1}{\sqrt{x}}>0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>0\end{cases}}\Leftrightarrow x>1\)

2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}< 0\end{cases}}\)( dễ thấy trường hợp này không xảy ra :> )

Vậy với x > 1 thì A > 0

Giúp mình với mình đang cần gấp. Thk you các pạn

P = \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\). \(\frac{\left(x-1\right)^2}{2}\)( x\(\ge0\); x\(\ne\)1)

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

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

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

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

= -x² + \(\sqrt{x}\)+ 2

b. tự tính nha

c, P = -x² + \(\sqrt{x}+2\) 

           =  - (x² - 2.x.1/2 + 1/4) +2 +1/4

          = - (x-1/2)²+ 9/4

          ta có  (x - 1/2)² \(\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\forall x\)

dấu "=" xảy ra khi và chỉ khi x-1/2 = 0

                                               x=1/2

vậy GTLN của P= 9/4 khi và chỉ khi x=1/2

#mã mã#

\(\frac{4+\sqrt{X}}{7}\)

Đề bài đâu bn ơi 

Nếu rút gọn thì mình làm cho

Ta có: \(P=\left(\frac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\frac{1-\sqrt{x}}{\sqrt{x}}+\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)\)         (    ĐKXĐ: \(x\ge1\))

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

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

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

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

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

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

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

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

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

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

P=\(x+2\sqrt{x}+1\)

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

Với lại ko nhân \(\sqrt{x}\)vào \(\sqrt{x}-1\)nha sr