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

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

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

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

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

\(=\sqrt{x}\)

b) Để P>4 thì \(\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Khi x>16 thì P>4

a) ĐKXĐ: \(x,y\ge0\)

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

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

b) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)

\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

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

\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)

a) ĐKXĐ: \(x\ge0;x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

Giúp mình với

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Thay x=0 vào A, ta được:

\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)

\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)

\(=\dfrac{11}{3}-2-1\)

\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)

a: ĐKXĐ: x>1; x<>2

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

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

c: Khi x=3+2căn 2 thì

P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1

\(a,dkxd:x\ge0,x\ne4\)

\(b,B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\dfrac{1}{\sqrt{x}-2}\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x^2}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(c,x=16\left(tm\right)\Rightarrow B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{4+2}{4\left(4-2\right)}=\dfrac{6}{8}=\dfrac{3}{4}\)

\(d,B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Leftrightarrow\sqrt{x}+2>0\Leftrightarrow\sqrt{x}>-2\left(ktm\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

Kết hợp với \(dk:x\ge0\) ta kết luận \(0\le x< 4\) thì \(B>0\).

a) Điều kiện xác định:

\(\left\{{}\begin{matrix}x-2\sqrt{x}\ne0\\x\ge0\end{matrix}\right.\)\(\Leftrightarrow x>0,x\ne4\)

Vậy...

b) \(B=\dfrac{\sqrt{x}.\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{1}{\sqrt{x}-2}\)

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

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

c) Tại x=16 ( thỏa mãn đk) thay vào B đã rút gọn ta được:

\(B=\dfrac{\sqrt{16}+2}{\sqrt{16}\left(\sqrt{16}-2\right)}=\dfrac{3}{4}\)

d) \(B>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)

\(\Leftrightarrow\sqrt{x}-2>0\)\(\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

Vậy x>4 thì B>0

`a)ĐKXĐ:{(x > 0),(x \ne 4):}`

`b)` Với `x > 0,x \ne 4` có:

`A=[\sqrt{x}(\sqrt{x}+2)+\sqrt{x}(\sqrt{x}-2)]/[x-4].[x-4]/[\sqrt{4x}]`

`A=[x-2\sqrt{x}+x-2\sqrt{x}]/[2\sqrt{x}]`

`A=[2\sqrt{x}(\sqrt{x}-2)]/[2\sqrt{x}]=\sqrt{x}-2`

`c)` Với `x > 0,x \ne 4` có:

`A < 3 <=>\sqrt{x}-2 < 3<=>\sqrt{x} < 5<=>x < 25`

           Kết hợp đk

 `=>0 < x < 25 ,x \ne 4`