【1】9×10^2010-10^2011
=9×10^2010-10×10^2010
=-10^2010
【2】9³-9²-8×9²
=9²(9-1-8)
=0
【3】（2x+3)²-2（2x+3)(3x-2)+(3x-2)²
=(2x+3+3x-2)²
=(5x+1)²
【4】2（x-1)(x²+x+1)+6(x+1)²=32+2（x+1)（x²+2x+1)
2x³-2+6(x+1)²=32+2(x³+1)+2(x+1)×3x
6x²+12x+6=32+4+6x²+6x
6x=30
x=5
【5】求证：对于任意整数n,（2n+1)²-1一定能被8整除.
(2n+1)²-1
=(2n+1+1)(2n+1-1)
=4n(n+1)
∵n(n+1)能被2整除
∴4n(n+1)能被8整除
（2n+1)²-1一定能被8整除
【6】设n为整数,求证（2n+1)²-25能被4整除
（2n+1)²-25
=(2n+1+5)(2n+1-5)
=4(n+3)(n-2)
∴（2n+1)²-25
【7】如果2x-y=10,求代数式[(x²+y²）-（x-y)²+2y(x-y)]÷4y的值
[(x²+y²）-（x-y)²+2y(x-y)]÷4y
=(x²+y²-x²+2xy-y²+2xy-2y²)÷4y
=(4xy-2y²)÷4y
=(2x-y)/2
=10/2
=5
【8】已知x(x-1)-(x²-y)=2,求（x²+y²）÷2-xy的值
x(x-1)-(x²-y)=2
x²-x-x²+y=2
-x+y=2
x-y=-2
（x²+y²）÷2-xy
=(x²-2xy+y²)/2
=(x-y)²/2
=4/2
=2
【9】已知a-b-c=4,求1/2a(a-b-c)+b(1/2×c-1/2×a+1/2×b）+1/2×c（b+c-a)的值
1/2表示二分之一
1/2a(a-b-c)+b(1/2×c-1/2×a+1/2×b）+1/2×c（b+c-a)
=1/2a(a-b-c)-1/2b(a-b-c)-1/2c(a-b-c)
=1/2(a-b-c)²
=1/2×16
=8
【10】已知a=1/2×m+1,b=1/2×m+2,c=1/2×m+3,求a²-ab+b²-ac-bc+c²的值
a²-ab+b²-ac-bc+c²
=1/2[(a-b)²+(b-c)²+(c-a)²]
=1/2[(-1)²+(-1)²+2²]
=3
