Lời giải:

$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$

$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$

$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$

\(P=\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

\(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}-2}{x-4}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)

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

Với \(x\ge0;x\ne4\)

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

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

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

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

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

b) ĐKXĐ : \(x\ne\pm1\)

\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

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

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

c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

a) ĐKXĐ : \(x\ge0;x\ne16\)

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

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)

\(=\left(\dfrac{\sqrt{x}.\left(\sqrt{x}-4\right)}{x-4}+\dfrac{4.\left(\sqrt{x}+4\right)}{x-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\left(\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\right).\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{x+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

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

\(=\dfrac{3x+9\sqrt{x}+4x-12\sqrt{x}-7x+3}{x-9}:\dfrac{2\sqrt{x}-4-\sqrt{x}+3}{\sqrt{x}-3}\)

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

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

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

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

=>x<1

Sửa đề: 

ĐKXĐ x\(\ge0,x\ne1,x\ne4\)

(x+2√x+4x√x−8+x+2√x+4x−1):(3+1√x−2+2√x+1)

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

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

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

P=\(\dfrac{x\sqrt{x}+x-9}{3\left(x-3\right)}\)

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

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

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

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

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

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