1.\(\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

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

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

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

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

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =-\sqrt{3}\)

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

\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)

\(\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)

\(a,=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\\ b,=\dfrac{x^2+4x+3+x^2+4x+4}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{x+1}{x+3}\left(x\ne-1;x\ne-2;x\ne-3\right)\\ =\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)

\(a,\dfrac{4}{\sqrt{x}+1}+\dfrac{2}{1-\sqrt{x}}-\dfrac{\sqrt{x}-5}{x-1}\)

\(=\dfrac{4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{4\sqrt{x}-4-2\sqrt{x}-2-\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

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

\(b,\left(\dfrac{x+1}{x+2}+\dfrac{x+2}{x+3}\right):\dfrac{x+3}{x+1}\)

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

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

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

\(=\dfrac{2x^2+8x+7}{\left(x+2\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)

\(=\dfrac{\left(2x^2+8x+7\right)\left(x+1\right)}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{\left(2x^2+8x+7\right).x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{2x^3+8x^2+7x+2x^2+8x+7}{\left(x+2\right)\left(x+3\right)^2}\)

\(=\dfrac{2x^3+10x^2+15x+7}{\left(x+2\right)\left(x+3\right)^2}\)

a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)

=>\(4⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

=>\(x\in\left\{16;4;25;1;49\right\}\)

b: 

Bài 1:
a. ĐKXĐ: $3x\geq 0$

$\Leftrightarrow x\geq 0$

b. ĐKXĐ: $\frac{x-1}{x+3}\geq 0$

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x-1\geq 0\\ x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} x-1\leq 0\\ x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 1\\ x< -3\end{matrix}\right.\)

Bài 2:

\(C=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{2+2\sqrt{2.3}+3}-\sqrt{2-2\sqrt{2.3}+3}\)

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

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

\(=2\sqrt{2}\)

Bài 3:
a. 

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

\(=\frac{2}{\sqrt{x}+2}:\frac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{2}{\sqrt{x}+2}.\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}}=\frac{2(\sqrt{x}-2)}{\sqrt{x}}\)

b. Khi $x=\frac{1}{4}$ thì $\sqrt{x}=\frac{1}{2}$.

Khi đó $A=\frac{2(\frac{1}{2}-2)}{\frac{1}{2}}=-6$

c.

$A=\frac{2(\sqrt{x}-2)}{\sqrt{x}}=2-\frac{4}{\sqrt{x}}$

$< 2$ do $\frac{4}{\sqrt{x}}>0$

Ta có đpcm

d. Với $x$ nguyên, để $A$ nguyên thì $\sqrt{x}$ là ước của $4$

$\Leftrightarrow \sqrt{x}\in\left\{1;2;4\right\}$

$\Rightarrow x\in\left\{1;4;16\right\}$ (đều tm)

b) lấy kết quả rút gọn của câu A ta được

 \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}< 1.=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)

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

đề bài cho x>=0 ta suy ra luôn

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

vậy x <1 thì P < 1

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

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

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

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

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

Bài `1`

\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)

2:

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b: B=5

=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)

=>\(5\sqrt{x}+15=\sqrt{x}+8\)

=>\(4\sqrt{x}=-7\)(loại)

Vậy: \(x\in\varnothing\)