a) 

= \(\sqrt{18-6\sqrt{6}+3}\)        

= \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)       

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

= \(|3\sqrt{2}-\sqrt{3}|\)   

= \(3\sqrt{2}-\sqrt{3}\)   

b) 

= \(\sqrt{\frac{7}{2}-\sqrt{7}+\frac{1}{2}}\)   

= \(\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)    

= \(\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)     

= \(|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}|\) 

= \(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\)        

c) 

= \(\sqrt{3+2\sqrt{3}+1}\)  

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

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

d) 

Đặt t = \(\sqrt{x-1}\left(ĐK:t\ge0\right)\)   

= \(\sqrt{t^2+1-2t}\)       

= \(\sqrt{\left(t+1\right)^2}\)   

\(=t+1\)      

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

\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2\sqrt{9}\sqrt{6}+3}=\sqrt{\left(\sqrt{18}\right)^2-2\sqrt{18}\sqrt{3}+\left(\sqrt{3}\right)^2}\)

                                \(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}=\sqrt{3}+3\sqrt{2}\)

\(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)

                           \(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{2}}{2}\)

\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Với \(x\ge1\)thì \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)

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

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

T đã tốn mấy phút cuộc đời viết lời giải cho bạn r, tiếc j mấy giây mà bấm k cho t ik =))

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

Tương tự cho cái còn lại sẽ ra

Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)

\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)

Vậy A<B

k đi mình làm cho

Giả sử \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)\(\le\sqrt{3}\)

<=> 4 + \(\sqrt{7}\)+ 4 - \(\sqrt{7}\)- 2×\(\sqrt{16-7}\)\(\le3\)

<=> 8 - 6 \(\le3\)

<=> 2 \(\le3\)(đúng)

Vậy \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)< √3

\(\sqrt{4+7}-\sqrt{4-\sqrt{7}}=2,152902878\)

\(\sqrt{3}=1,732050808\)

Rùi so sánh đi