a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))

a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

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

\(=\sqrt{xy}\)

b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)

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

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

a: Khi x=25 thì \(A=\dfrac{5-2}{5-1}=\dfrac{3}{4}\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\)

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

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

\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}=\dfrac{x-4}{x-1}\)

c: \(P=\dfrac{A}{B}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

P<1/2

=>P-1/2<0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\)

=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(x\in\varnothing\)

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

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

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

a: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

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

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

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

c: A/B>4/3

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{4}{3}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{4}{3}>0\)

=>\(\dfrac{3\left(\sqrt{x}+1\right)-4\sqrt{x}}{3\sqrt{x}}>0\)

=>\(3\left(\sqrt{x}+1\right)-4\sqrt{x}>0\)

=>\(3\sqrt{x}+3-4\sqrt{x}>0\)

=>\(-\sqrt{x}>-3\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: 0<x<9

a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)

\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)

b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)

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

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

\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

c: P=A/B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

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

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

=>P<2

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

__________________________________________________________________________________________

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!

\(\sqrt{\left(120-11\right)^2}+\sqrt{\left(10-\sqrt{120}\right)^2}\)

\(=120-11+10+\sqrt{120}\)

\(=\sqrt{120}\left(\sqrt{120}+1\right)-1\)

\(a,=\left(120-11\right)+\left|10-\sqrt{120}\right|=109+\sqrt{120}-10=99+2\sqrt{30}\\ b,=\sqrt{\left(\sqrt{x+1}+1\right)^2-\left(\sqrt{x+1}+1\right)^2}=\sqrt{0}=0\)

a) Thay x=25 vào B ta có:

\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)

b) \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-1}{x-5\sqrt{x}+6}\)

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

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

\(A=\dfrac{x-9-x+4+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(A=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(A=\dfrac{2}{\sqrt{x}-2}\)

c) Ta có: \(A>B\) Khi:

\(\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{2-\sqrt{x}-2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow0< x< 4\)