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

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

a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)

b, với x > 0 

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

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

a) \(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{3.\left(\sqrt{x}-3\right)+x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-5-\left(\sqrt{x}-3\right)}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

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

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

b) \(M< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)

Kết hợp điều kiện ta được \(0< x< 4\) thì M < 0

c) Từ câu b ta có M < 0 \(\Leftrightarrow0< x< 4\)

nên \(x\inℤ\) để M nguyên âm <=> \(x\in\left\{1;2;3\right\}\)

Thay lần lượt các giá trị vào M được x = 1 thỏa 

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

Vì x > 4 nên \(\sqrt{x}-2>0\)

Áp dụng BĐT Cauchy ta có 

\(M=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\ge2\sqrt{\left(\sqrt{x}-2\right).\dfrac{4}{\sqrt{x}-2}}+4=8\)

Dấu "=" xảy ra khi \(\sqrt{x}-2=\dfrac{4}{\sqrt{x}-2}\Leftrightarrow x=16\left(tm\right)\)

1) \(M=\left(\dfrac{3}{\sqrt[]{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{x-3\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x}}\right)\left(x>0;x\ne9\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\left(\sqrt[]{x}-3\right)}{\left(\sqrt[]{x}+3\right)\left(\sqrt[]{x}-3\right)}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}-\dfrac{1}{\sqrt[]{x}}\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}-9+x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}+x}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\sqrt[]{x}+3}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+3\right)}{x-9}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)

\(\Leftrightarrow M=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}-2}\)

\(\Leftrightarrow M=\dfrac{x}{\sqrt[]{x}-2}\)

2) Để \(M< 0\) khi và chỉ chi

\(M=\dfrac{x}{\sqrt[]{x}-2}< 0\left(1\right)\)

Nghiệm của tử là \(x=0\)

Nghiệm của mẫu \(\sqrt[]{x}-2=0\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)

Lập bảng xét dấu... ta được

\(\left(1\right)\Leftrightarrow0< x< 4\)

3) \(M=\dfrac{x}{\sqrt[]{x}-2}\inℤ^-\)

\(\Leftrightarrow x⋮\sqrt[]{x}-2\)

\(\Leftrightarrow x-\sqrt[]{x}\left(\sqrt[]{x}-2\right)⋮\sqrt[]{x}-2\)

\(\Leftrightarrow x-x+2\sqrt[]{x}⋮\sqrt[]{x}-2\)

\(\Leftrightarrow2\sqrt[]{x}⋮\sqrt[]{x}-2\)

\(\Leftrightarrow2\sqrt[]{x}-2\left(\sqrt[]{x}-2\right)⋮\sqrt[]{x}-2\)

\(\Leftrightarrow2\sqrt[]{x}-2\sqrt[]{x}+4⋮\sqrt[]{x}-2\)

\(\Leftrightarrow4⋮\sqrt[]{x}-2\)

\(\Leftrightarrow\sqrt[]{x}-2\in\left\{-1;-2;-4\right\}\)

\(\Leftrightarrow x\in\left\{1;0\right\}\)

\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)

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

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

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

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

Với `x>0` có:

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

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

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

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

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

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

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

\(=1\)

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

A= (( (x+3).\(\sqrt{x}\) +4x)/(x+9).\(\sqrt{x}\) +6x):((x\(\sqrt{x}\) +x)/x\(\sqrt{x}\) +3x)

A= (( (x+3).\(\sqrt{x}\) +4x/(x+9).\(\sqrt{x}\) +6x)):((\(\sqrt{x}\) +1)/\(\sqrt{x}\) +3)

A= (( (x+3).\(\sqrt{x}\) +4x/(x+9).\(\sqrt{x}\) +6x)).(​(\(\sqrt{x}\) +3)/​\(\sqrt{x}\)+1)​

A= (x^2 + 3x+4x\(\sqrt{x}\) + 3x\(\sqrt{x}\) + 9\(\sqrt{x}\) +12x)/x^2 + 9x + 6x\(\sqrt{x}\) + x\(\sqrt{x}\) +9\(\sqrt{x}\) +6x

A= (x^2 + 15x + 9\(\sqrt{x}\) +7x\(\sqrt{x}\))/x^2 + 15x +7x\(\sqrt{x}\) + 9\(\sqrt{x}\) 

A=1 

Lời giải:
ĐK: $x\geq 0; x\neq 4; x\neq 9$

\(P=\frac{1}{\sqrt{x}+1}:\left[\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}\right]\)

\(=\frac{1}{\sqrt{x}+1}:\frac{x-9-(x-4)+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

Để $P>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+1}>0$

$\Leftrightarrow \sqrt{x}-2>0$ (do $\sqrt{x}+1>0$)

$\Leftrightarrow x>4$

Kết hợp với ĐKXĐ suy ra $x>4; x\neq 9$

a, \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

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

\(P=\dfrac{1}{\sqrt{x}+1}:\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

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