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: \(=\sqrt{2}-1\)

b: \(=\sqrt{3}+1+2-\sqrt{3}=3\)

Cái này chắc rút gọn :

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

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

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

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

a) \(1=\sqrt{1}< \sqrt{2}\)

b) \(2=\sqrt{4}>\sqrt{3}\)

c) \(6=\sqrt{36}< \sqrt{41}\)

d) \(7=\sqrt{49}>\sqrt{47}\)

e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)

f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)

g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)

h) \(\sqrt{3}>0>-\sqrt{12}\)

i) \(5=\sqrt{25}< \sqrt{29}\)

\(\Rightarrow-5>-\sqrt{29}\)

\(B=10+5\sqrt{3}+10-5\sqrt{3}=20\)

\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)

\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)

\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)

\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)

\(=6+\sqrt{5}\)

\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)

\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)

\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)

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

\(=-3+\sqrt{3}+\sqrt{5}\)

\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)

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

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

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

\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

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

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

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

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

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

\(=\sqrt{6}\)

\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)

Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)

Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

.... Đúng thì ủng hộ nha ....
 Kết bạn với mình ... ;) ;)

đặt \(x=\frac{\sqrt{3}}{cost};\forall t\in\left(0;\frac{\pi}{2}\right)\Rightarrow tant>0\)

\(dx=d\left(\frac{\sqrt{3}}{cost}\right)=\frac{-\sqrt{3}sint}{cos^2t}dt\)

Thay vào, ta có \(\int\frac{\sqrt{3}\cdot\frac{-\sqrt{3}sint}{cos^2t}}{\frac{\sqrt{3}}{cost}\sqrt{\frac{3}{cos^2t}-3}}dt=\int\frac{-3\cdot\frac{sint}{cos^2t}}{\frac{3}{cost}\cdot\sqrt{tan^2t}}dt=\int\frac{-sint}{cost\cdot tant}dt=-\int dt=-t+C\)

Bây giờ thay t vào là ra