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I got 2268a base 11 digits... I'm not sure how you calculated it but I used Melody's answer as a guide and this is what I got? 

11^(8^5) in base 11 ==1.1062AA481 E+33957

It has: 33,957 + 1 ==33, 958 digits in base 11

1 - Arithmetic mean =[3 + 147] / 2 ==75

2 - Geometric mean=Sqrt[3 x 147]=sqrt(441) == 21

Bruh I know you are a alt for clairepow STOP cheating on AOPS. This will not be my last warning.

Please try and refrain from giving full answers. 

Sometimes is ok but many people are here solely to cheat.  Clair is definitely one of those people.

Excuse the handwriting :)

Excuse the handwriting :)

Correct thedudemanguyperson...!

\(\frac{1}{\sqrt{2}+\frac{1}{\sqrt{8}+\sqrt{200}+\frac{1}{\sqrt{18}}}}\)

                                        \(​​​​\frac{1}{\sqrt{8}+\sqrt{200}+\frac{1}{\sqrt{18}}}=\frac{3\sqrt{2}}{73} \)

\(=\frac{1}{\sqrt{2}+\frac{3\sqrt{2}}{73}}\)

                                        \(\sqrt{2}+\frac{3\sqrt{2}}{73}:{\frac{76\sqrt{2}}{73}}\)                  

\(=\frac{1}{\frac{76\sqrt{2}}{73}}\)

                                      \(\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}\)

\(=\frac{73}{76\sqrt{2}}\)

                                      \(Rationalize \frac{73}{76\sqrt{2}}:\frac{73\sqrt{2}}{152}\)

\(=\frac{73\sqrt{2}}{152}\)

This question answered, 2 years ago here is the link https://web2.0calc.com/questions/help_41734

thats what i got 

If we define the smallest integer to be \(x\), then

\(x + (x+5) + (x+7) = 42.\)

\(3x + 12 = 42\)

The integers are x, x + 5, x + 7, or 10, 15, 17.

Logarhythm's answer is correct.

Oh, both of these answers or just mine?

Hmmm, I got something similar to what Log got...

\(\frac{y-2}{y^2\:-\:9}\:-\:\frac{y-7}{y^2\:+\:4y\:-\:21}\)

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{y^2+4y-21}\)

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{\left(y-3\right)\left(y+7\right)}\)

\(=\frac{\left(y-2\right)\left(y+7\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}-\frac{\left(y-7\right)\left(y+3\right)}{\left(y-3\right)\left(y+7\right)\left(y+3\right)}\)

\(=\frac{\left(y-2\right)\left(y+7\right)-\left(y-7\right)\left(y+3\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

\(=\frac{9y+7}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

Excuse the handwriting..

Nice answer Cubey, but I think I got something different from yours..

\(\frac{y-2}{y^2 - 9} - \frac{y-7}{y^2 + 4y - 21}\)

\(\frac{y-2}{(y+3)(y-3)} - \frac{y-7}{(y+7)(y-3)}\)

\(\frac{(y+7)(y-2) - (y+3)(y-7)}{(y+3)(y-3)(y+7)}\)

\(\frac{(y^2 + 5y - 14) - (y^2 - 4y - 21)}{(y+3)(y-3)(y+7)}\)

\(\frac{y+7}{(y+3)(y-3)(y+7)}\)

\(\boxed{\frac{1}{(y+3)(y-3)}}\)

:0 I think this is for spirit of math right

4b² - 12b = 4b(b - 3)

27b + 18 = 9(3b + 2)

18b³ - 128b = 2b(9b² - 64)

                    = 2b(3b + 8)(3b - 8)

63b² - 282b - 216 = 3(21b² - 94b - 72)

                             = 3(3b + 2)(7b - 36)

Then you put these over the fractions:

= 4b(b - 3) / 2b(3b + 8)(3b - 8)  -   9(3b + 2) / 3(3b + 2)(7b - 36)

= 2(b - 3) / (3b + 8)(3b - 8)  -  3 / (7b - 36)

= (2(b - 3)(7b - 36)  -  3(3b + 8)(3b - 8)) / (3b + 8)(3b - 8)(7b - 36)

= (14b² - 114b + 216 - 27b² + 192) / (3b + 8)(3b - 8)(7b - 36)

= (-13b² - 114b + 408) / (3b + 8)(3b - 8)(7b - 36)   ; b ≠ 0 , ±8/3 ,  -2/3 , 36/7

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

\(\frac{2 - 4\sqrt{3}}{13} + \frac{6 + 3\sqrt{3}}{1}\)

\(\frac{(2 - 4\sqrt{3} )+ 13(6 + 3\sqrt{3})}{13}\)

\(\frac{80 +35\sqrt{3}}{13}\)

They satisfy $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_1z_2$, so done by https://math.stackexchange.com/questions/953129/equilateral-triangle-from-three-complex-points

Taking a b out of the numerator and denominator of the first  one, we get $\frac{4b-12}{18b^2-128}=\frac{2b-6}{9b^2-64}=\frac{2b-6}{(3b+8)(3b-8)}$.

FOr the second, the denominator factors as $3(3b+2)(7b-6)$, and $27b+18=9(3b+2)$. You can do the rest, I believe in you!!