a - a always equal 0

The correct answer is C

We have: 

\(a^2+b^2+c^2\ge ab+bc+ca\)

The "=" happen when and only when a = b = c

Applying the Cauchy-Schwarz inequality, we have:

\(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\le\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2+2+2}\le\dfrac{9}{ab+bc+ca+6}=\dfrac{9}{9}=1\)

So......................

\(\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{\dfrac{17}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{4}{17}}}=\dfrac{1}{4+\dfrac{1}{\dfrac{72}{17}}}=\dfrac{1}{4+\dfrac{17}{72}}=\dfrac{1}{\dfrac{305}{72}}=\dfrac{72}{305}\)

\(\left(\left(X^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}=\sqrt{\sqrt{\sqrt{X}}}=\sqrt[2^3]{X}=\sqrt[8]{X}\)

The fraction in odd location have the denominator is sequence of odd numbers from 3

The fraction in even location have the denominator is sequence of 2ⁿ from 2¹

And their numerator is 1

So the fraction in ? is \(\dfrac{1}{13}\)

Their average is: \(\dfrac{6+7+42}{16}=\dfrac{55}{16}=3,4375\)

\(\dfrac{30\%\cdot\dfrac{25}{17}}{\dfrac{3}{4}\cdot\dfrac{1}{2}}=\dfrac{\dfrac{15}{34}}{\dfrac{3}{8}}=\dfrac{20}{17}\)

200 days = 28 weeks and 4 days

So 200 days from now will be 3 + 4 = 7 or Saturday 

Call a zeeko is x, a teeko is y. We have:

\(\dfrac{1}{2}x=60\%y\)

\(\Leftrightarrow0,5x=0,6y\)

\(\Leftrightarrow x=1,2y\)

\(\Leftrightarrow\dfrac{x}{y}=1,2\)

With this post, we have:

130%A = B

60%B = C

120%C = D

=> D = 60% of 120% of B

=> D = 72%B

\(\dfrac{1}{10\sqrt{10}}=\dfrac{1}{\sqrt{10^3}}\)

\(\dfrac{\sqrt{10}}{\sqrt{10^4}}=\dfrac{1}{\sqrt{10^3}}\)

So A = B

The answer is c

\((x-1)(2x-1) = 9-x\)

=> \(2x^2 - 3x + 1 = 9- x\)

=> \(2x^2 - 2x = 8\)

=> \(2x^2 - 2x - 8 = 0\)

=> \(2(x^2 - x - 4)= 0\)

=> \(2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{17}{2}=0\)

=> \(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{2}\)

=> \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)

=> \(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

3rd term is: 1 + 2 = 3

4th term is: 1 + 2 + 3 = 6

5th term is: 1 + 2 + 3 + 6 = 12

....

15th term is: 1 + 2 + 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = 1536 

\(0.2\overline{36}=\dfrac{236-2}{990}=\dfrac{234}{990}=\dfrac{13}{55}\)

Mr. Bee 

People call this triangle is Pascal.

                                                  1

                                        1                    1

                              1                    2                    1

                    1                    3                    3                    1

          1                    4                    6                    4                    1

1                    5                    10                    10                    5                    1

So the answer is 1,5,10,10,5,1

This sequence is ...,5k,6k,7k,8k,9k,10k,11k,12k,... with k = 4

So each terms in this will \(⋮4\)

So in 4 answers, answer d is correct :]

This sequence is 0k,1k,2k,3k,4k,5k,6k,7k,8k,...,300k with k = 3

So the missing numbers in "..." is 9k to 299k

So there are: 299 - 9 + 1 = 291 numbers in this

\(S_{ABC}=\dfrac{\left(13-11\right)\left(17-2\right)}{2}=\dfrac{2\cdot15}{2}=15\left(unit^2\right)\)