a: Ta có: \(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}\)

\(=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)

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

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

b) Ta có: \(\sqrt{\left(1-\sqrt{2006}\right)^2}\cdot\sqrt{2007+2\sqrt{2006}}\)

\(=\left(\sqrt{2006}-1\right)\left(\sqrt{2006}+1\right)\)

=2005

a) đk: \(\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)

Ta có:
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)

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

\(A=\frac{4\sqrt{a}+4a\sqrt{a}-4\sqrt{a}}{a-1}\cdot\frac{a+1}{\sqrt{a}}\)

\(A=\frac{4a\left(a+1\right)}{a-1}\)

b) Ta có: \(a=\sqrt{4+\sqrt{15}}\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4^2-\sqrt{15}^2}\)

\(=\sqrt{10}-\sqrt{6}\)

\(\Rightarrow A=\frac{4\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{10}-\sqrt{6}+1\right)}{\sqrt{10}-\sqrt{6}-1}=...\)

1. Ta có 4=2 căn 4 

Căn 4<căn 5

=> 2 căn 5 >4

2. Ta có 3^2=9 =16-7=16-căn 49

( căn 15 -1)^2

= 15 -2 căn 15 +1= 16-2 căn 15 =16- căn 60

Căn 60>căn49

=> 3> căn 15 -1

3. Ta có  6^2=36=27+9= 27+ căn 81

    (căn 26 +1)^2=26 +2 căn 26 +1=27+ 2 căn 26 =27+ căn 52

 Căn 52< căn 81 

=> 6> căn 26+1

4. Ta có (căn 2 -2)^2 =2- 4 căn 2+4=6- 4 căn 2

             (căn 3 -3 )^2 = 3 -6 căn 3 +9= 12- 6 căn 3

      Lại có 8 căn 2 =căn 128

                6 căn 3 =căn 108

=> (căn 3 -3)^2> 2(căn 2 -2)^2 

=> căn 3 -3 > căn 2-2 

\(2\sqrt{5}>4\)

\(3< \sqrt{15-1}\)

\(6>\sqrt{26-1}\)

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

a/ \(\left(\sqrt{18}\right)^2-2\cdot\sqrt{18}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{18}-\sqrt{3}\right)^2\)

b/\(\left(\sqrt{54}\right)^2-2\cdot\sqrt{54}+1=\left(\sqrt{54}-1\right)^2\)

c/\(\left(\sqrt{9}\right)^2-2\cdot\sqrt{9}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{9}-\sqrt{5}\right)^2\)

d/\(\left(\sqrt{8}\right)^2+2\cdot\sqrt{8}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{8}+\sqrt{5}\right)^2\)

Ta có: \(\sqrt{7}< \sqrt{9}\); \(\sqrt{15}< \sqrt{16}\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

 Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề hơn nhé. 

a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)