a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

a,Ta có  \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)

\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)

b, Với \(x\ge0;x\ne1\)

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

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

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

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

Vậy biểu thức ko phụ thuộc biến x 

c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên 

thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

a. Ta có $x=4-2\sqrt{3}=1-\mathrm{2.1.}\sqrt{3}+3=(1-\sqrt{3}{)}^2$.

$\Rightarrow \sqrt{x}=|1-\sqrt{3}|=\sqrt{3}-1.$

1x−1=13−1−1=13−2=−2−3.

b. Với $x\ge 0$ và $x\ne 1$ ta có:

−3xxx−1−11−x):(1−x+21+x+x)

−3x(x−1)(x+x+1)+1x−1]:(1+x+x)−(x+2)1+x+x

−3x+x+x+1(x−1)(x+x+1).x+x+1x−1

x−2x+1(x−1)(x+x+1).x+x+1x−1 

a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

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

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

Bài 1:

ĐKXĐ: \(\dfrac{5}{x^2+6}>=0\)

=>\(x^2+6>0\)

mà \(x^2+6>=6>0\forall x\)

nên \(x\in R\)

Bài 2:

a: Sửa đề: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\cdot\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)

b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

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

\(=\dfrac{\sqrt{12}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\)

e: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=\sqrt[3]{-27}=-3\)

f: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

\(=\sqrt{2}+1-\sqrt{2}+1=2\)

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)

a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)

Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)

b, Với \(x\ge0;x\ne4\)

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

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

c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay 

\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)

\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)

\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))

sao cô cho cả đáp án ra lun thế ạ @@

à ko em nhầm nhầm em xin lỗi cô 

1) A = 2 khi x =9

3) 2 + \(\sqrt{x}\)- 3x = 0

-3x +\(\sqrt{x}\)+2 = 0

3x- \(\sqrt{x}\)- 2 =0

Đặt \(\sqrt{x}\)= t ; ĐK : t\(\ge\)0; t \(\ne\) 2

\(^{3t^2-t-2=0}\)

t1 =1 (TMĐK)

t2 = \(\dfrac{-2}{3}\)(Không TMĐK)

Với t = 1 => \(\sqrt{x}=1\Leftrightarrow x=1\)