1²+(1×2）²+2²=9=3² 2²+(2×3）²+3²=49=7² 3²+(3×4）²+4²=169=13² What rule do you find? Please use the equation containing n (n is a positive integer) and explain the reason

The rule of [answer] is as follows: the rule: n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\178; = on the right the law holds

1) From (x + a) (x + b) = x ^ 2 + (a + b) x + AB, we can get x ^ 2 + (a + b) x + AB = (x + a) (x + b), that is, the quadratic trinomial x ^ 2 + (a + b) x + AB about X can be decomposed into the product of two quadratic factors about X: x ^ 2 + (a + b) x + AB = (x + a) (x + b)
Comparing their coefficients, what rules do you find? Can you factorize the polynomial x ^ 2-5x + 6 with the rules you find?
(2) By observing the equation (x + 3) (2x + 1) = 2x ^ 2 + 7x + 3, (x + 1) (2x + 3) = 2x ^ 2 + 5x + 3, we can get 2x ^ 2 + 7x + 3 = (x + 3) (2x + 1), 2X ^ 2 + 5x + 3 = (x + 1) (2x + 3)
Comparing their coefficients, what law do you find? Can you factorize the polynomial 3x ^ 2-8x + 4 with the law you find?

Cross method
Just mention it:
1:
1 -2
1 -3
1 * -3 + 1*-2 = -5
So: (X-2) (x-3)
2:
1 -1
3 -4
1 * -4 + 3 * -1 = -8
(x-1)(3x-4)

Observe the following
1^2+(1X2)^2+2^2=9=3^2
2^2+(2X3)^2+3^2=49=7^2
3^2+(3X4)^2+4^2=169=13^2
What rule do you find? Please use the equation containing n (n is a positive integer) and explain the reason

n^2+[n(n+1)]^2+(n+1)^2=(n^2+n+1)^2

(2a+b)(2a-3b)+a(2a+b)
I want to know the process, the more detailed the better, don't just answer, I will do it later

The original formula = (2a + b) (2a-3b + a) extracts the common factor (2a + b)
=(2a + b) (3a-3b) extract the common factor 3
=3(2a+b)(a-b)

Bracket 5 / 8 minus 1 / 6 minus 1 / 3 includes the process of multiplying by 24

Given the circle C: (x + 1) 2 + y2 = 25 and point a (1, 0), q is a point on the circle, and the vertical bisector of AQ intersects CQ at m, then the trajectory equation of point m is______ ．

From the equation of the circle, we can see that the center of the circle is C (- 1,0), the radius is equal to 5, let the coordinates of the point m be (x, y), ∵ AQ's vertical bisector intersects CQ at m, | Ma | = | MQ |, and | MQ | + | MC | = radius 5, | MC | + | Ma | = 5 ＞| AC |. According to the definition of ellipse, the locus of point m is an ellipse with a and C as the focus, and 2A = 5, C = 1, | B = 212, so the elliptic equation is x2254 + & nbsp; y2214 = 1, that is 4x225 + 4y221 = 1, so the answer is 4x225 + 4y221 = 1

Bracket 2.25 plus (3 / 5-1.21 × 11 / 5) bracket divided by 2 / 5, need formula

Bracket 2.25 plus (3.5-1.21 × 11 / 5) bracket divided by 2 / 5! Bracket 2.25 plus (3.5-1.21 × 11 / 5) bracket divided by 2 / 5 = bracket 2.25 plus (3.5-1.21 × 11 / 5) bracket multiplied by 5 / 2 = bracket 2.25 plus (18-20 / 3)

Simple calculation 1999 * 0.5 / 1997 * 0.3 + 1999 / 1.2

1. More than ten times more than ten: pithy formula: head multiply head, tail add tail, tail multiply tail. For example: 12 × 14 =? 1 × 1 = 12 + 4 = 62 × 4 = 812 × 14 = 168 note: if the number of bits is multiplied, 0 should be used for less than two digits. 2

A simple method to calculate (1) 2 ^ 10-2 ^ 9-2 ^ 8 (2) shows that 3 ^ 200-4 × 3 ^ 199 + 10 × 3 ^ 198 can be divided by 7

(1)2^10-2^9-2^8=2^8(4-2-1)=2^8=256
(2)3^200－4×3^199＋10×3^198=3^198(9-12+10)=3^198 *7
So 3 ^ 200-4 × 3 ^ 199 + 10 × 3 ^ 198 contains a factor of 7, so it can be divided by 7

(x + 5) square + 3 (x + 5) - 4 = 0 with factorization, 9 (x-1) square = (2x + 1) square with factorization

（x+5)²+3(x+5)-4=0
(x+5+4)(x+5-1)=0
x=-9 x=-4
9（x-1)²=(2x+1)²
9（x-1)²-(2x+1)²=0
[3（x-1)+(2x+1)][3（x-1)-(2x+1)]=0
x=2/5 x=5

1²+(1×2）²+2²=9=3² 2²+(2×3）²+3²=49=7² 3²+(3×4）²+4²=169=13² What rule do you find? Please use the equation containing n (n is a positive integer) and explain the reason