Known X1 1 X2 2 X3 3 丨X20 x2013 2013 0 Try to find the value of the algebraic formula 2x1 2x2 2x3 2x

Because |x1-1|+|x2-2|+|x3-3|+…丨x2012-2012丨+|x2013-2013|=0 each term has an absolute value, so each term is greater than or equal to 0, and they add up to =0, so x1-1=0, x2-2=0......x2013-2013=0, x1=1, x2=2,...x2013=2013, so algebraic.

2x1-2x2-2x3-……2x2012-2x2013

2^1-2^2-2^3-……2 2012-2 2013 (2 1 means 2 of the first of 2, 2 2013 of 2 of 2).

Finally, use the first n terms of the proportional series and the formula to put (2 1 + 2 2 + 2 3 + ......2 2012+2 2013) calculated, the formula: sn=[a1x(1-q n)] (1-q),a1=2,q=2,n=2013 you bring it in yourself, and then subtract the result with 4, and it comes out, I hope it can help you)

And also. Landlord, you made a mistake in the question supplement, and the required formula is all minus: 2x1-2x2-2x3-......2x2012-2x2013 instead of 2x1-2x2-2x3 -......2x2012+2x2013

Since each absolute value is greater than the equal inclusion of the field accompanied by zero, the sum of these absolute values can only be 0, so x1=1, x2=2, so 2 x1-2 x2-2=2-4-2=-4

Solution: After dividing the two sides of the equation by the regret ruler x deformation, the value of x 2 + [1 x 2 can be obtained by squared on both sides;

According to the value of x 2 + 1 x 2, the value of x + 1 x] is obtained by using the perfect square formula, and the formula is simplified by the square difference formula, and the value can be obtained by substituting the respective values into the calculation

x2-3x-1=0，x-

1 x]=3, squared on both sides: (x-[1 x])2=x2+[1

x2-2=9, then x2+

x2=11；

x+1/x]）2=x2+[1

x2+2=13，x+

1/x]=±

13, then x2-[1

x2=（x+

1/x]）（x-[1/x]）=3

Comments: Test points for this question: Perfect square formula; Square Difference Formula

Test Center Comments: This question examines the perfect square formula, the square difference formula, and mastering the perfect square formula is the key to solving this question 1 year ago.

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Probably similar issues.

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Solution: To consider the non-negativity of absolute values, it is necessary to make |x1-1|+|x2-2|+|x3-3|+…x2013-2013|=0, required.

x1=1x2=2

x2013=2013

Solution: |x1-1|+|x2-2|+|x3-3|+…x2002-2002|+|x2003-2003|=0，x1=1，x2=2，x3=3，…，x2002=2002，x2003=2003，2x1-2x2-…-2x2002+2x2003=2-22-…-22002+22003

6 Therefore, the algebraic formula 2x1-2x2 -...-2x2002+2x2003 has a value of 6

Since each absolute value is greater than or equal to zero, now the sum of these absolute values is 0, and each absolute value can only be 0, so x1 = 1, x2 = 2, so 2 x1-2 x2-2 = 2-4-2 = -4

The absolute value of a number must be greater than or equal to 0, and if so many absolute values in the known conditions add up to 0, they can only be 0 each.

So x1 = 1, x2 = 2, x3 = 3,......x2010=2010。

Therefore, the judgment is 2010-2009 + 2008-2007 + ......4-3+2-1

Two two for a group of subtraction, then it becomes a lot of 1 added up, and the total number of regret is 2010 2.

So the answer is 1005.

Arrange a set of monomials according to the rules - x2010 times, x2009 times, y, -x2008 times, x2007 times, y3 times,,, do 1, find what are the 99 of the first Senpeiheng??? 2. What is the 2010th one?

Because the absolute value is greater than or equal to 0, so that the sum of the 2010 absolute Bizen values is equal to 0, then all the absolute values are equal to 0, so there are a few quiets:

x1-1=0,x2-2=0,..x2010-2010=0x1=1,x2=2,..x2010=2010x2010-x2009+x2008-……x4-x3+x2-x1