Try to find the remainder of x 111 x 31 x 13 x 9 x 3 divided by x 1.

(x^111 + x^110) -x^110 + x^109) +x^109 + x^108) -

x^33 + x^32) -x^32 + x^31) +x^13 + x^9 - x^3)

In the above equation, all but the last three are already divisible by (x+1). Let's consider the last three next.

x^13 + x^9 - x^3

x^13 + x^12) -x^12 + x^11) +x^11 + x^10) -x^10 + x^9) +2x^9 - x^3)

In the above equation, all but the last term are already divisible by (x+1). Let's consider the latter two items next.

2x^9 - x^3

2(x^9 + x^8) -2(x^8 + x^7) +2(x^5 + x^4) -2(x^4 + x^3) +x^3

In the above equation, all but the last term are already divisible by (x+1). Let's consider the last item.

x^3 (x^3 + x^2) -x^2 + x) +x + 1) -1

Therefore, the final covariance is -1

x^(n+1)+x^n=x^n(x+1)

x+1)|(x^(n+1)+x^n)

That is, make up x (n+1) + x n shape.

Original = (x 111 + x 110) - (x 110 + x 109) + (x 109 + x 108).

Remainder x 13 + x 9-x 3

x^13+x^12)-(x^12+x^11)+(x^11+x^10)-(x^10+x^9)+2x^9-x^3

Remaining 2x 9-x 3

2(x^9+x^8)-2(x^8+x^7)+.2(x^5+x^4)-2(x^4+x^3)+x^3

surplus x 3 (x 3 + x 2) - (x 2 + x) + (x + 1) - 1 surplus - 1

Because the equivalents must be constants, the result of substituting x=-1 is that the odd exponent does not matter.

x 111 - x 31 + x 13 + x 9 - x 3 x 31 * (x 80-1) + x 3 (x 10-1) + x 9 because.

x^80-1=(x-1)(x^79+x^78+..x 1+1) x 10-1 is also divisible by the anterior band.

So left x 9

x^9-1+1=(x-1)(x^8+x^7+..1) +1 repents of the lack of 1 in the fierce reed

x 285 x 83 x 71 x 9 x 3 divided by x 1

Solution 1: x 285 x 83 x 71 x 9 x 3 3 (x 285 1) (x 83 1) (x 71 1) (x 9 1) (x 3 1) 4

Because x 285 1, x 83 1, x 71 1, x 9 1, x 3 1 are divisible by x 1.

So, the remainder of the original formula divided by x 1 is 4

Solution 2 is governed by the remainder theorem, where the remainder is equal to x 285 x 83 x 71 x 9 x 3 3 at x 1, i.e.

Remainder 1285 183 171 19 13 3 4

Comment: Among the two solutions to this problem, solution 1 is to decompose the part of the original formula that can be divisible by x 1 through identity deformation, and the rest is the remainder Solution 2 is to find the remainder through the remainder theorem, which is the general method of this kind of problem, and it is necessary to be proficient

Let f(x)=x243

x81+x27

x9+x3x=q(x)(x-1)+r, then f(1)=q(1) 0+r=r, i.e.: r=f(1)=1243

So the answer is: 6

Summary. 3x + (11-x) = 29 solutions.

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Because it is two absolute values.

Therefore, the two sides can be squared at the same time to get x -18x+81=x -26x+169, and the transfer of the term will be defeated and the same kind of term can be combined.

That is, there is (26-18) x = 169-81 = 88, and the solution is x=88 8 = 11