12+56+1112+1920+2930+… +97019702+98999900．

12+56+1112+1920+2930+… +97019702+98999900．

12+56+1112+1920+2930+… +97019702+98999900=（1-12）+（1-16）+（1-112）+… +（1-19900）=1×99-（12+16+112+… +19900），=99-（11×2+12×3+13×4… +199×100），=99-（1-12+12-13+13−14+… +199-1100），=99-（1-1100），=99-99100，=981100．

About a math problem, how much is 20 / 29 + 3 / 20?
What is 20 out of 29 + 3 out of 20?
What is 20 out of 29 + 3 out of 20?
What is 20 out of 29 + 3 out of 20?
What is 20 out of 29 + 3 out of 20?
It's better to explain why

20 of 29 + 3 of 20 = 400 / 580 + 87 / 580 = 487 / 580

How many degrees is 32 degrees 59 points equal to?

The conversion between minutes and degrees is 60, so 32 degrees and 59 minutes = 32.983 degrees, which circulates at 3

A mathematical problem in square ABCD, triangle BCE is equilateral triangle, proof: angle ead equal to angle EDA equal to 15 degrees
In square ABCD, triangle BCE is equilateral triangle. Prove that angle ead equals angle EDA equals 15 degrees

Because the angle Abe = angle ECD = 30 degrees, ab = DC, EB = EC,
So, the triangle Abe is equal to the triangle DCE, AE = De,
Angle ead is equal to angle EDA
Let the side length of the square be 1, let EF be parallel to ab through e, let BC be crossed to F, and let ad be extended reversely to g,
EF = (radical 3) / 2
Eg = 1 - (radical 3) / 2
Tangent value of angle ead = eg / (AD / 2) = (1 - (radical 3) / 2) / (1 / 2) = 2-radical 3
tg30=tg(2*15)=2tg15/(1-tg15*tg15)
The tg15 = 2-radical 3 is substituted into the above formula, and the result is proved to be true

Given that f (x) = 6x + 7 / 2x-2013 ①, find out f (1) + F (2012), f (2) + (2011) ②
It is known that f (x) = 6x + 7 / 2x-2013
① Find out f (1) + F (2012), f (2) + (2011)
② Come up with a common conclusion from the result of (1) and prove it
③ Find the value of F (1) + F (2) + (3) + ········· + F (2010) + (2011) + F (2012)