We know that there is no number whose square is negative in the range of rational numbers. Now let's make a sharp intellectual turn. Suppose that the square of a number I is exactly equal to - 1. Under this assumption, complete the following questions: I + I ^ 2 + I ^ 3 + I ^ 4 = I + I ^ 2 + I ^ 3 + I ^ 4 +... + I ^ 2012=

We know that there is no number whose square is negative in the range of rational numbers. Now let's make a sharp intellectual turn. Suppose that the square of a number I is exactly equal to - 1. Under this assumption, complete the following questions: I + I ^ 2 + I ^ 3 + I ^ 4 = I + I ^ 2 + I ^ 3 + I ^ 4 +... + I ^ 2012=

If I ^ 2 = - 1, then I ^ 3 = I ^ 2 · I = - I, I ^ 4 = I ^ 2 · I ^ 2 = (- 1) × (- 1) = 1
∴i＋i^2＋i^3＋i^4＝i－1－i＋1＝0
And I ^ 5 + I ^ 6 + I ^ 7 + I ^ 8 = I ^ 4 · (I + I ^ 2 + I ^ 3 + I ^ 4) = 1 × 0 = 0
i^9＋i^10＋i^11＋i^12＝i^8·(i＋i^2＋i^3＋i^4)＝1×0＝0
…
i^2009＋i^2010＋i^2011＋i^2012＝i^2008·(i＋i^2＋i^3＋i^4)＝1×0＝0
So the original formula is 0

It is known that the product of three rational numbers a, B and C is negative, and their sum is positive. When x = ial / A + B / | B | + | C | / C, find the square of 2013x-2012x
Add the value of 2014

If the product of three is negative, it means that all three are negative or two are positive and one is negative
But a > 0 will get | a | / a = 1, a

It is known that a, B and C are three rational numbers which are not equal to zero, the product of which is negative, and the sum of which is positive

According to the meaning of the title, we can draw a conclusion
abc＜0,a+b+c＞0
A, B, C are all negative numbers, or two positive and one negative
If they are all negative numbers, then a + B + C < 0
Of the three numbers, two are positive and one is negative
Because these three numbers are symmetrical
If a > 0, b > 0 and C < 0 are not set, then
A / A + B / B + C / C|
=a/a +b/b +c/(-c)
=1+1-1
=1