And (1995X square+1996Y square+1997Z square) =1995cube root+1996cube root, find 65438.

A: XYZ is greater than 0, which means that all three are greater than 0 or one of them is greater than 0. According to the previous conditions, it is not possible for one of the three to be greater than 0, only if all three are greater than 0. Order 1995X cubic = 1996 y cubic = 1997 z cubic = k, then (1995X square+1996Y square) =

Cube root of 1995+cube root of1996+cube root of1997 = cube root of k/x 3+cube root of k/y 3 = k.

Then the cube root of (1/x+1/y+1/Z) = (1/x+1/z), and then (1/z).

2. As we all know:

6/((n+ 1)(n+2)(n+3)(n+3))=(a/(n+ 1))+(b/(n+2))(c/(n+3))(d/(n+4))

Where a, b, c and d are constants, and the value of a+2b+3c+4d is _ _ _ _ _ _.

A: 6/[(n+1) (n+2) (n+3) (n+4)]

= 6/{[(n+ 1)(n+4)][(n+2)(n+3)]}

=6/[(n squared +5n+4)(n squared +5n+6)

= 3/(square of n+5n+4)-3/(square of n+5n+6)

= 3/[(n+ 1)(n+4)]-3/[(n+2)(n+3)]

=[ 1/(n+ 1)- 1/(n+4)]-[3/(n+2)-3/(n+3)]

= 1/(n+ 1)+(-3)/(n+2)+3/(n+3)+(- 1)/(n+4)

So: a= 1 b=-3 c=3 d=- 1.

So: a+2b+3c+4d =1+2 * (-3)+3 * 3+4 * (-1) = 0.

3. As we all know, the following equation applies to any real number x (n is a positive integer):

( 1+x)+( 1+x)^2+( 1+x)^3…+( 1+x)^n=a0+a 1x+a2(x^2)…an(x^n)

And a 1+a2+a3...+an = 57, then the possible value of n satisfying the condition is _ _ _ _ _.

""stands for power, and the number after it is exponent.

0, 1, 2 … n on the right side of the equation are subscripts.

A: Solution:

(1) Let x=0.

Then there are: (1+0)+(1+0) 2+(1+0) 3+...+(1+0) n = A0+a1(0).

Namely:1+1+1+1...+1(n1) = A0+0.

Get n = a0.

(2) Let x= 1.

There are: (1+1)+(1+1) 2+(1+1) 3+(1+).

Namely: 2+2 2+2 3+...+2 n = A0+a1+A2+A3+...+An.

=2+2^2+2^3+………+2^n=+57

=2+2^2+2^3+………+2^n-57=

(2+2 2+2 3+...+2 n should have a formula, but I don't know, I can only use the assumed value of n to find it. Please forgive me! )

(3) Let n=6.

Then: 2+22+23+24+25+26-57 = n.

=2+4+8+ 16+32+64-57= n

= 126-57 = n n=69 and n=6 are contradictory.

∴n>6

Let n=5.

Then it is: 2+22+263+24+25-57 = n.

=2+4+8+ 16+32-57= n

= 62-57 = n n = 5 ∴ n = 5 hold.

∴n=5

4.