Put \(A=\dfrac{1}{5}+\dfrac{1}{13}+..........+\dfrac{1}{n^2+\left(n+1\right)^2}\)

We have:\(\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{2n^2+2n+1}=\dfrac{1}{2}\left(\dfrac{1}{n^2+n+\dfrac{1}{2}}\right)< \dfrac{1}{2}\left(\dfrac{1}{n^2+n}\right)=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

Apply we have:A<\(\dfrac{1}{2}\left(1-\dfrac{1}{2}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+............+\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+........+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

=\(\dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)< \dfrac{1}{2}\)

We have:A-1=\(\dfrac{3-4x}{x^2+1}-1=\dfrac{3-4x-\left(x^2+1\right)}{x^2+1}=\dfrac{3-4x-x^2-1}{x^2+1}=\dfrac{-\left(x^2+4x-2\right)}{x^2+1}=\dfrac{-\left(x+2\right)^2+6}{x^2+1}=\dfrac{-\left(x+2\right)^2}{x^2+1}+\dfrac{6}{x^2+1}\) Because:\(\dfrac{-\left(x+2\right)^2}{x^2+1}\le0\) so A-1\(\le\dfrac{6}{x^2+1}\) 

\(\Rightarrow\)max_A-1=\(\dfrac{6}{x^2+1}\)\(\Leftrightarrow x=-2\Rightarrow\)max_A-1=\(\dfrac{6}{5}\)\(\Rightarrow\)max_A=\(\dfrac{11}{5}\)

We have:9*9*9*........*9                      (345 number 9)

=9³⁴⁵=9³⁴⁴ * 9=(9²)¹⁷² * 9=81¹⁷² * 9=..............1 * 9=...............9

Answer:the last number of 9*9*9*...........*9 is:9

Put A=\(\dfrac{1}{3}+\dfrac{2}{3^2}+............+\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+............+\dfrac{100}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\dfrac{2}{3}+.........+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..........+\dfrac{100}{3^{100}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}< 1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}\)

\(\Rightarrow6A< 3+1+\dfrac{1}{3}+........+\dfrac{1}{3^{98}}\)

\(\Rightarrow6A-2A< \left(3+1+...........+\dfrac{1}{3^{98}}\right)-\left(1+\dfrac{1}{3}+........+\dfrac{1}{3^{99}}\right)=3-\dfrac{1}{3^{99}}\)

\(\Rightarrow4A< 3-\dfrac{1}{3^{99}}< 3\Rightarrow A< \dfrac{3}{4}\)

We have:A+1=\(x+\dfrac{1}{x}+1=\dfrac{x^2+x+1}{x}=0\)

\(\Rightarrow A+1=0\Rightarrow A=-1\)

Answer:A=-1

The area of the square triangle is:

     \(\dfrac{12.5}{2}=30\)

The area of the semicircle is:

      \(\dfrac{\dfrac{13}{2}.\dfrac{13}{2}.3,14}{2}=\dfrac{26533}{400}=66,3325\)

The total area of the shaded regions is:

       66,3325-30=36,3325\(\approx36\)

Answer:36

We have:\(\dfrac{1}{27}=0,\left(037\right)\)

Because after the decimal point we have three number:0,3,7 repeat and 41:3 residual 2 so the 41st digit after the decimal point in the decimal expansion of \(\dfrac{1}{27}\) is the second of three numbers here is 3

Answer:3

If a=12,b=4,c=5 and x=\(\dfrac{1}{2}\) so \(\dfrac{\left(\dfrac{abc}{x}\right)-\left(6b^2-4\right)}{0,5}\)\(=\dfrac{\left(\dfrac{12.4.5}{\dfrac{1}{2}}\right)-\left(6.4^2-4\right)}{0,5}\)

\(=\dfrac{\left(\dfrac{240}{\dfrac{1}{2}}\right)-\left(96-4\right)}{0,5}=\dfrac{480-92}{0,5}=\dfrac{388}{\dfrac{1}{2}}=388.2=776\)

Answer:776`

We have:\(\dfrac{x^4.\left(x^2\right)^5}{x^5}=xy\Leftrightarrow\dfrac{x^4.x^{10}}{x^5}=xy\Leftrightarrow x^9=xy\)

With x=0 so y\(\in\)Z

With x\(\ne\)0,because \(x^9=xy\Leftrightarrow x^8=y\)

Change: 1 hour = 3600 s

The number of feet he will travel in an hour is:

         44 x 3600=158400(ft)

The number of miles he will travel in an hour is:

          158400:5280=30(miles)

Answer:30 miles

Because \(0^{4^{100}}=0\) so \(2^{0^{4^{100}}}=2^0=1\)                                                          (1)

Other way:\(0^{100}=0\) so \(4^{0^{100}}=4^0=1\) so \(2^{4^{0^{100}}}=2^1=2\)                                 (2)

From (1) and (2) we have th value of the following expression is:\(2^{0^{4^{100}}}-2^{4^{0^{100}}}=1-2=-1\)

Alchemy I can't answer my question and I'm lazy

Alchemy I have a answer quicker than your answer but I want to a answer quickest.

Alchemy Your answer is right but I need a compact answer

We have:99999 x 55555

=100000 x 55555 - 55555

=5555500000-55555

=5555444445

Answer:99999 times 55555 is:5555444445

The sentences 1:       If |x+y|>|x|+|y|\(\Rightarrow\left(\left|x+y\right|\right)^2>\left(\left|x\right|+\left|y\right|\right)^2\Leftrightarrow x^2+2xy+y^2>x^2+2.\left|x\right|.\left|y\right|+y^2\)

\(\Rightarrow2xy>2\left|x\right|.\left|y\right|\Leftrightarrow xy>\left|x\right|.\left|y\right|\) (vô lí)\(\Rightarrow\left|x+y\right|\le\left|x\right|+\left|y\right|\)

The sentences 2:       If |x-y|<|x|-|y|\(\Rightarrow\left(\left|x-y\right|\right)^2< \left(\left|x\right|-\left|y\right|\right)^2\Leftrightarrow x^2-2xy+y^2< x^2-2.\left|x\right|.\left|y\right|+y^2\)

\(\Rightarrow-2xy< -2.\left|x\right|.\left|y\right|\Leftrightarrow2xy>2.\left|x\right|.\left|y\right|\)\(\Leftrightarrow xy>\left|x\right|.\left|y\right|\)(vô lí) \(\Rightarrow\left|x-y\right|\ge\left|x\right|-\left|y\right|\)

We have:\(x:y:z=a:b:c\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)

\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2}=\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}=\dfrac{1^2}{1}=1\)

\(\Rightarrowđpcm\)

Because 2,3,1,7>0 so 2 isn't the second number

We can put off:4 x 4=16(number)

But 2 isn't the second number so we can put off:16-4=12(number)

Because 2+1=3\(⋮3\) ; 2+7=9\(⋮3\) so 21 and 27 isn't a prime number

So we can put off:12-2=10(number) but 33 and 77 isn't a prime number

Answer:10-2=8(number)

Answer:\(\dfrac{\dfrac{\dfrac{\dfrac{\dfrac{123,456-6}{10}-5}{10}-4}{10}-3}{10}-2}{10}\)

=\(\dfrac{\dfrac{\dfrac{\dfrac{\dfrac{7341}{625}-5}{10}-4}{10}-3}{10}-2}{10}\)=\(\dfrac{\dfrac{\dfrac{\dfrac{2108}{3125}-4}{10}-3}{10}-2}{10}=\dfrac{\dfrac{\dfrac{-5196}{15625}-3}{10}-2}{10}\)

\(=\dfrac{-0,3332544-2}{10}=-0,223332544\)

Put n=2k because 2 | n

We have:2^2k-1=4^k-1

Because 4\(\equiv1\left(mod3\right)\)\(\Rightarrow4^k\equiv1^k=1\left(mod3\right)\Rightarrow4^k-1⋮3\)

\(\Rightarrow2^n-1\) is a multiple of 3 then 2 | n