Exploratory question X 1 X 1 X Square 1

1, Write 2 to the 6th power + 2 to the 5th power + 2 to the 4th power + 2 to the 3rd power + 2 + 1 to the 6th power of 2 + the 5th power of 2 + the 4th power of 2 + the 3rd power of 2 + 2 to the 3rd power + 2 + 1 + 4-4 (2 to the 6th power + 2 to the 5th power + 2 to the 4th power + 2 to the 3rd power + 2 to the 2nd power + 2 + 1) -4 This is the 6th power of 2 -4 2, the same way...

2 to the 6th power + 2 to the 5th power + 2 to the 4th power + 2 to the 3rd power + 2 to the square + 2 + 1 equals. 2 to the 6th power + 2 to the 5th power + 2 to the 4th power + 2 to the 3rd power + 2 to the square + 2 + 1) (2-1) (2-1).

Amount. 2 to the 7th power of -1) (2-1) is equal to.

The solution to the second question is the same as above.

2-1) (2 to the 6th power + 2 to the 5th power + 2 to the 4th power + 2 to the 3rd power + 2 to the 2nd power + 2 + 1) = 2 to the 7th power of 2 - 1 = 127

2-1) (2 to the power of 2005 + 2 to the power of 2004 + 2 to the power of 2003 + ....+2+1)=2006 to the power of -1

2 to the power of 1 = 2

2 to the 5th power = 32

2 to the power of 2 = 4

2 to the power of 6 = 64

2 to the 3rd power = 8

2 to the power of 7 = 128

2 to the power of 4 = 16

2 to the 8th power = 256

2006 4 surplus 2

So the end number is 4-1=3

1 + 2 + 2 squared + 2 to the 3rd power + ...2 to the 63rd power = (2 to the 64th power - 1) divided by (2-1) = 2 to the 64th power - 1

The single digit is 7

The original formula <=> (Note: each time the first two items are extracted 1+x common factor) (1+x)(1+x)+x(1+x) 2+....+x(1+x)^2012

1+x）^2·（1+x）+.x(1+x)^2012=...

1+x）^2012·（1+x）

1+x）^2013

It's not hard to understand. ~

1) Lift the ant to clear the cave and take the common factor with the positive grinding 2 times.

2) Method 2004 times Result: (1+x) Stuffy 2005 power.

Solution: (1) The above epochal chain method of factoring is the common factor method, which is applied twice

2) The above method needs to be applied 2004 times, and the basal pulse result is (1+x) 2005

3) Solution: original formula = (1+x)[1+x+x(x+1)]+x(x+1)2+....+x（x+1）n，（1+x）2（1+x）+x（x+1）2+…+x(x+1)bounce n,(1+x)3+x(x+1)3+....+x（x+1）n，（x+1）n+x（x+1）n，（x+1）n+1．

Let (x-1) = t original formula = t2-t-6 = 0 multiply the cross to get t = 3 or -2

Let t=x-1 give x=-1 or 4

Hello: x 1) squared (x 1) 6 0

x-1-3)(x-1+2)=0

x-4)(x+1)=0

x1=4 x2=-1

If there are other questions about the limb buried in the liquid file, click to ask me for help, it is not easy to answer the question, please understand, thank you.

Good luck with your studies! o(oThank you.)

(x-1)(x+1)(x²+1)

x²-1)(x²+1)

x 4-1 x 4 is X's fourth world-famous sock.

Solution: (1).

x-1)(x⁶+x⁵+x⁴+x³+x²+x+1)=x⁷-1(2)(x-1)(xⁿ+xⁿ⁻¹x+1)=xⁿ⁺¹1(3)1+2+2²+.2³⁴+2³⁵=(2-1)(2³⁵+2³⁴+

Problem solving idea (3): Notice that the formula is multiplied by 1, the value is unchanged, and 2-1=1, so multiply the formula by 2-1 to obtain the formula summarized in the previous law, where x=2, and then the result is obtained.