Giải

Ta có:

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

Khi đó:

\(x^2=\left(\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\right)^2\\ =2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\\ =8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-3\left(2+\sqrt{3}\right)}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{6-3\sqrt{3}}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-\sqrt{2}.\sqrt{12-6\sqrt{3}}\\ =8-\sqrt{2}.\left(\sqrt{4+2\sqrt{3}}+\sqrt{12-6\sqrt{3}}\right)\\ =8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}+\sqrt{9-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\right)\\ 8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\right)\\ =8-\sqrt{2}.\left(\sqrt{3}+1+3-\sqrt{3}\right)\\ =8-4\sqrt{2}\\ \Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16.\left(8-4\sqrt{2}\right)\\ =96-64\sqrt{2}-128+64\sqrt{2}=-32\)

Vậy \(S=-32\)

\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~ 

\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

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

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

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

\(=\)\(1\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 sai thì thông cảm >.< 

\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)

\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì  \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)

\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)

= \(\sqrt{6}\)

a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)

d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)

e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

\(\dfrac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}.\sqrt{5}+2\sqrt{5}-3\sqrt{2}.\sqrt{3}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

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

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

Bạn coi lại xem dưới mẫu đúng dấu ''+'' không á, phải dấu ''-'' mới rút với tử ở trên được nha.

1/ Bạn trên làm rồi mình không làm lại.

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

\(=\frac{3\sqrt{2}+3\sqrt{3}-3\sqrt{5}+\sqrt{10}+\sqrt{15}-5}{2\sqrt{6}}+\frac{3\sqrt{2}-3\sqrt{3}+3\sqrt{5}-\sqrt{10}+\sqrt{15}-5}{-2\sqrt{6}}\)

\(=\frac{3\sqrt{2}+3\sqrt{3}-3\sqrt{5}+\sqrt{10}+\sqrt{15}-5-3\sqrt{2}+3\sqrt{3}-3\sqrt{5}+\sqrt{10}-\sqrt{15}+5}{2\sqrt{6}}\)

\(=\frac{6\sqrt{3}-6\sqrt{5}+2\sqrt{10}}{2\sqrt{6}}=\frac{3}{\sqrt{2}}-\frac{3\sqrt{5}}{\sqrt{6}}+\frac{\sqrt{5}}{\sqrt{3}}=\frac{9\sqrt{2}-3\sqrt{30}+2\sqrt{15}}{6}\)

\(\frac{x^2-2x+2007}{2007x^2}=\frac{x^2}{2007x^2}-\frac{2x}{2007x^2}+\frac{2007}{2007x^2}=\frac{1}{2007}-\frac{2}{2007x}+\frac{1}{x^2}\)

đặt t = 1/x

=> \(\frac{1}{2007}-\frac{2}{2007x}+\frac{1}{x^2}=\frac{1}{2007}-\frac{2t}{2007}+t^2=\frac{1}{2007}-\frac{2t}{2007}+\frac{2007t^2}{2007}=\frac{2007t^2-2t+1}{2007}\)

giải theo kiểu casio 570 VN PLUS cho nhanh nhé

bấm MODE 5 3 2007 = -2 = 1 = = = = =

ra gtnn của 2007t² - 2t + 1 là 2006/2007 tại t = 1/2007

vậy gtnn của \(\frac{2007t^2-2t+1}{2007}\)là \(\frac{\frac{2006}{2007}}{2007}\)tại t = 1/2007

t = 1/2007  => 1/x = 1//2007  => x = 2007

vậy x = 2007 thì biểu thức có gtnn

Cái này chắc rút gọn :

\(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{2.2+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}\)

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

\(\sqrt{12+2\sqrt{35}}=\sqrt{7+2\sqrt{7.5}+5}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}+\sqrt{5}\)

\(\sqrt{12+2\sqrt{35}}=\sqrt{5}+\sqrt{7}\)

\(\sqrt{12+2\sqrt{35}}=\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}=\left|\sqrt{5}+\sqrt{7}\right|=\sqrt{5}+\sqrt{7}\)