\(đkxđ\Leftrightarrow x\ge4\)

\(P=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)

\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\sqrt{\frac{4^2}{x^2}-2.\frac{4}{x}+1}}\)

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

\(=\frac{|x-2|+|x-6|}{|\frac{4}{x}-1|}=\frac{x-2+|x-6|}{|\frac{4}{x}-1|}\)

Dùng bảng xét dấu nha

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

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

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

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

Đề hơi sai sai ý bạn ơi

\(a.P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{10\sqrt{x}-2x}\left(x>0,x\ne4,x\ne25\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right].\dfrac{x-4}{10\sqrt{x}-2x}\)

\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{10\sqrt{x}-2x}\)

\(=\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}\left(5-\sqrt{x}\right)}\)

\(=\dfrac{\sqrt{x}}{5-\sqrt{x}}\)

\(b.\) Thay \(x=\dfrac{1}{4}\) vào P, ta được:

\(\dfrac{\sqrt{\dfrac{1}{4}}}{5-\sqrt{\dfrac{1}{4}}}=\dfrac{0,5}{5-0,5}=\dfrac{1}{9}\)

Vậy ......................

\(c.P< -1\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{5-\sqrt{x}}< -1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+5-\sqrt{x}}{5-\sqrt{x}}< 0\)

\(\Leftrightarrow\dfrac{5}{5-\sqrt{x}}< 0\)

\(\Leftrightarrow5-\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}>5\)

\(\Leftrightarrow x>25\left(tm\right)\)

Vậy ...................

\(a,P=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}-2x-4\sqrt{x}+8}{6\sqrt{x}-18}\left(dk:x\ne4,x\ge0,x\ne9\right)\)

\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2}\right).\dfrac{\sqrt{x^2}\left(\sqrt{x}-2\right)-4\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}.\dfrac{\left(x-4\right)\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2-x-3\sqrt{x}-2}{\left(x-4\right)\left(\sqrt{x}-2\right)}.\dfrac{\left(x-4\right)\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)

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

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

\(b,P>0\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-3}>0\Leftrightarrow-\sqrt{x}>0\Leftrightarrow\sqrt{x}< -1\left(ktm\right)\)

\(\Leftrightarrow\sqrt{x}-3>0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

\(c,P< 1\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-3}< 1\Leftrightarrow-\sqrt{x}< 1\Leftrightarrow\sqrt{x}>-1\left(ktm\right)\)

\(\Leftrightarrow\sqrt{x}-3< 1\Leftrightarrow\sqrt{x}< 4\Leftrightarrow x< 2\)

a: \(P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)-2\sqrt{x}\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-3\right)}\)

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

=1/3(căn x-3)

b: P>0

=>căn x-3>0

=>x>9

c: P<1

=>P-1<0

=>\(\dfrac{1-3\sqrt{x}+9}{3\sqrt{x}-9}< 0\)

=>\(\dfrac{-3\sqrt{x}+10}{3\sqrt{x}-9}< 0\)

=>(3căn x-10)/(3căn x-9)>0

=>x>100/3 hoặc 0<x<9

Điều kiện: x>2

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

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

P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b) P= \(\dfrac{1}{4}\)

⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)

⇔\(4\sqrt{x}-8=3\sqrt{x}\)

⇔\(\sqrt{x}=8\)

⇔x=64 (TM) 

Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)