Olympiad Solving Factoring Pay attention to the detailed solution, I m just a junior high school stu

Question 1: To examine the construction of a perfect level, we need to decompose +3 into +4 and -1.

Solution: a 2-b 2+4a + 2b + 3

a 2+4a+4)-(b 2-2b+1) (This step puts the latter term with a minus sign).

a+2) 2-(b-1) 2 I forgot what it was called, a 2-b 2=(a+b)(a-b))).

a+2+b-1)(a+2-b+1)

a+b+1)(a-b+3)

Scold. The process is as follows:2 2006+(-2) 2007 P.S

2) 2007=2 2006 (-2) The original formula can be reduced to: 2 2006+2 2006 (-2) Extract the common factor: 2 2006 (1-2).

After the original decomposition, it is: -2 2006

It should be tied up. I don't know if you have an answer, and the solution is -2...

It's a pattern finding question.

It can be deduced from the exponent of the factor.

x+1）＾ⁿ

Here's what it turned out of.

It is derived from the relationship between its exponents for the first time.

According to the law.

The second is (1+x) to the 6th power.

The last one is (n+1) power of (1+x).

Solution: 1, from x y xy 2x y 1 0 (x+1) 2+y 2=y(x+1)>=0, the deformation has (x+1-y) 2=-y(x+1)>=0, so y(x+1)<=0, and y(x+1)>=0, so y(x+1)=0, then (x+1) 2+y 2=0, there is y=0, x=-1

2 3 4 1 25 5 =(2 2+1) 2,2 3 4 5 1 121 11 =(3 2+2) 2,3 4 5 6 1 361 19 =(4 2+3) 2, from which we can see that the law is n (n+1) (n+2) (n+3)+1=((n+1) 2+n) 2 (where n>=1 and is an integer), which is expressed in mathematical language as the product of four consecutive natural numbers (the first natural number is greater than or equal to 1) and the sum of 1 is equal to the square of the second natural number and the square of the sum of the first natural number.

The equation of 1 can be reduced as: (x+1) +y(y-x-1)=0 x, where y is a real number.

Then there is only :(x+1) =0 , and y(y-x-1)=0 gives x=-1,y=0,-2

2 On the left is the multiplication of 4 consecutive natural numbers plus 1

(2nd power of x-3) + (2nd power of x -3)-2 This problem uses the pq formula (x+p)(x+q)=x 2+(p+q)x+pq

If the commutation method is used to make the 2nd power of x -3 to k, then the original formula becomes: k 2+k-2=(k-1)(k+2).

Swap k back to the 2nd power of x-3 and it becomes {(2nd power of x-3)+2}{(2nd power of x-3)-1}=(2nd power of x-1)(2nd power of x-4) = (x-1)(x+1)(x-2)(x+2)(x+2), got it? I answered so carefully, let's give some points.

Solution: Because x 3+y 3+z 3-3xyz=(x+y+z)(x 2+y 2+z 2-xy-yz-zx).

So (x 3+y 3+z 3-3xyz) (x+y+z)=x 2+y 2+z 2-xy-yz-zx

So x 3+y 3+z 3-3xyz is divisible by (x+y+z).