（急!证明对于任意自然数n,3^(n+2) - 2^(n+3)+3^n-2^(n+1)一定能被10整除.（1）证明对于任意自然数n,3^(n+2) - 2^(n+3) + 3^n - 2^(n+1)一定能被10整除.（2）若a-b=2,a-c=0.5,求（b-c）^2 - 3(b-c)+9／4的值.（3）已知6x^2-5xy+y^2+mx+ny+2=(3x-y-2)(2x-y-1).求证：m=-7,n=3.

(1)(n+2) - 2^(n+3) + 3^n - 2^(n+1)
=3^n*9-2^n*8+3^n-2^n+2
=3^n(9+1)-2^n(8+2)
=10(3^n-2^n)
(2)(a-c)-(a-b)=b-c=-1.5
（b-c）^2 - 3(b-c)+9／4=(b-c-3/2)^2=0
(3)已知6x^2-5xy+y^2+mx+ny+2=(3x-y-2)(2x-y-1) (3x-y-2)(2x-y-1)
=6x^2-5xy+y^2-7x+3y+2
m=-7 ,n=3

（1）原式=10（3^n-2^n）
(2)原式=(b-c-3/2)^2=9
(3)右边=6x^2-5xy+y^2-7x+3y+2

3^(n+2) - 2^(n+3) + 3^n - 2^(n+1)
=(3^2*3^n+3^n)-(2^3*2^n+2*2^n)
=3^n(3^2+1)-2^n(2^3+2)
=10*3^n-10*2^n
=10*(3^n-2^n)
所以一定能被10整除
a-b=2,a-c=0.5
相减
(a-b)-(a-c)=2-0.5
c-b=1.5
b-c=-1.5
（b-c）^2 - 3(b-c)+9／4
=2.25+4.5+2.25
=9
(3x-y-2)(2x-y-1).
=6x^2-3xy-3x-2xy+y^2+y-4x+2y+2
=6x^2-5xy+y^2-7x+3y+2
=6x^2-5xy+y^2+mx+ny+2
对应项系数相等
所以m=-7,n=3

(1)3^(n+2) - 2^(n+3) + 3^n - 2^(n+1)
=3^(n+2)+3^n-(2^(n+3)+2^(n+1))
=3^n(1+3^2)-2^(n+1)(1+2^2)
=3^n*10-2^(n+1)*5
=10(3^n-2^n)
所以原式一定能被10整除
(2)b-c=0.5-2=-1.5
原式=（b-c-3/2）^2=(-1.5-1.5)^2=9
(3)将右边化开，得(3x-y-2)(2x-y-1)=6x^2-5xy+y^2-7x+3y+2
与左边对应，立得m=-7,n=3