Simplify the addition and subtraction of integers (1) (n + 2 power of X - N + 1 power of 3x - n power of 2x) - 2 (n power of X - N plus 2 power of 5x) (2) The nth power of a B - [the nth power of 2A + (the nth power of ab-the nth power of 2B)] - 2 (the nth power of B-the nth power of a)

Simplify the addition and subtraction of integers (1) (n + 2 power of X - N + 1 power of 3x - n power of 2x) - 2 (n power of X - N plus 2 power of 5x) (2) The nth power of a B - [the nth power of 2A + (the nth power of ab-the nth power of 2B)] - 2 (the nth power of B-the nth power of a)

（x^(n+2)-3x^(n+1)-2x^n）-2（x^n-5x^(n+2)）=x^(n+2)-3x^(n+1)-2x^n-2x^n+10x^(n+2)=11x^(n+2)-3x^(n+1)-4x^na^n*b-[2a^n+（ab^n-2b^n）]-2（b^n-a^n）=a^n*b-2a^n-ab^n+2b^n-2b^n+a^n=a^n*b-a^n-ab^n

A and B are shopping in the mall. First, a goes to the third floor underground to park, while B is on the second floor. Their colleagues walk up the stairs. It is known that when a goes to the fifth floor, B goes to the third floor. Q: what floor will a and B meet

Difference between a and B 3 + 2-1 = 4 floors
4÷（5-3）=2
Meet at 2 + 3x2 = 8

Mathematics problem of grade one in junior high school
1. 1 times 2 x plus 2 times 3 X. + 1999 times 2000 x = 1999
2. A △ B = ab-b + 1 is defined to solve the equation (3 △ y) △ 4 = 73
3. Let ABCD be four different natural numbers, and a × B × C × d = 2002, find the maximum possible value of a + B + C + D
I'll get more points if it's OK!! >_

X * (1 / 1 * 2 + 1 / 2 * 3 + 1 / 3 * 4 + ······ + 1 / 1999 * 2000 = 1999 (then split term)
&Nbsp; & nbsp; & nbsp; X * (1-1 / 2 + 1 / 2-1 / 3 + 1 / 3-1 / 4 + ····· + 1 / 1999-1 / 1999 + 1 / 2000 = 1999)
       x*（1-1/2000)=1999
       1999/2000*x=1999
                                x=1999*1999/2000
                               x=2000
2.    (3y-y+1)△4=73
                (2y+1)△4=73
                 8y+4-4+1=73
                            8y=72
                              y=9
3. 2002 = 2 * 7 * 11 * 13 * 1 & nbsp; & nbsp; maximum = 11 * 13 + 1 + 7 + 2 = 153

If the sum of the fifth power of MX × the N + 1 power of Y and the fourth power of 2 / 3 × x × y (where m is the coefficient) is equal to 0, then M=______ ,a=______ ,n=______ .

mx^5×y^(n+1)+2/3×x^a×y^4=0
Then the two terms are of the same kind and the coefficients are opposite
So m = - 2 / 3
5=a
n+1=4
So m = - 2 / 3, a = 5, n = 3