a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

d) Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

Đề đúng nhỉ ,bạn xem lại đề dùm mình ạ \(\sqrt[]{}\)X -2 hay là \(\sqrt[]{^{ }}\)x-2 

\(a,DKXD:x\ge0\)

\(b,A=\sqrt{x-\sqrt{x^2-4x+4}}\)

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

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

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

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

\(=\sqrt{2}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

b) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

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

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)

a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\)      (ĐK : \(\forall x\in R\))

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

     * Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)

     *Nếu x<2   => M=2-x-x-2=-2x

b,Để M=2\(\ne-4\)

     =>M=-2x

    =>-2x=-4

    =>x=2

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

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

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

     * Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)

    * Nếu x<2  =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)

             VẬY.......

 Tk nha!

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-2>0\\x-4\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x>2\\x\ne4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

mik thấy đề sai sai

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{x-4}\right):\dfrac{\sqrt{x}+1}{x-4}\)

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

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

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\\rightarrow \left\{{}\begin{matrix}x>\sqrt{2}\\x>-\sqrt{2}\\x>0\end{matrix}\right.\\ \rightarrow x>\sqrt{2}\)

Vậy \(x>\sqrt{2}\)

b) 

\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\\ =\left[\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{2\sqrt{x}}=\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}=\sqrt{x}\)

Vậy \(M=\sqrt{x}\)

a) ĐKXĐ:

\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}x>4\\x>-4\\x>0\end{matrix}\right.\\ \rightarrow x>4\)

Vậy \(x>4\)

TXĐ \(\sqrt{x}\)lớn hơn hoặc bằng 0=>x lớn hơn hoặc bằng 0

A=\(\sqrt{x}\)-\(\sqrt{x^2-4x+4}\)=\(\sqrt{x}\)-\(\sqrt{\left(x-2\right)^2}\)=\(\sqrt{x}\)-x+2

A=-(x-\(\sqrt{x}\)-2)=-(\(\sqrt{x}\)-2)(\(\sqrt[]{x}\)+1)

\(Đk:x\ge0\)

b) \(\sqrt{x}-\sqrt{x^2-4x+4}\)

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

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

Th1 : \(x-2\ge0\)

PT ( 1 ) \(=\sqrt{x}-x+2\)

Th2 : \(x-2< 0\)

PT ( 1 ) \(=\sqrt{x}-2+x\)