Question

已知X(X+1)-(X^2+Y)=1求代数式1/2(X^2+Y^2)-XY

1 已知X(X+1)-(X^2+Y)=1求代数式1/2(X^2+Y^2)-XY

2 (1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)

PS:有过程

Answer 1

解：
1、
X(X+1)-(X^2+Y)
=X^2+X-X^2-Y
=X-Y
=1

(X^2+Y^2)/2-XY
=[(X-Y)^2+2XY]/2-XY
=(X-Y)^2/2+XY-XY
=(X-Y)^2/2
=1/2

2、设Y=(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
两边同乘(1-2)得
(1-2)Y=(1-2)(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
=(1-2^2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
=(1-2^4)(1+2^4)(1+2^8)(1+2^16)
=(1-2^8)(1+2^8)(1+2^16)
=(1-2^16)(1+2^16)
=1-2^32

即-Y=1-2^32
Y=2^32-1=4294967295

Answer 2

1 已知X(X+1)-(X^2+Y)=1求代数式1/2(X^2+Y^2)-XY

解:由x(x+1)-(x^2+y)=1得
x*x+x-x^2-y=1
x^2+x-x^2-y=1
x^2-x^2+x-y=1
x-y=1

1/2(X^2+Y^2)-XY
=(x^2+y^2)/2-xy
=1/2*(x^2+y^2-2xy)
=1/2*(x-y)^2
=1/2*1^2
=1/2

2 (1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
=-(1-2)(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
=-(1-2^2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
=-(1-2^4)(1+2^4)(1+2^8)(1+2^16)
=-(1-2^8)(1+2^8)(1+2^16)
=-(1-2^16)(1+2^16)
=-(1-2^32)
=2^32-1

Answer 3

1、由已知得：X-Y=1,即X=1+Y，代入代数式
原式=1/2[(1+Y)^2+Y^2]-(1+Y)Y
=1/2+Y^2+Y-Y-Y^2
=1/2
2、不好意思，不会！

Answer 4

1)y=x-1
1/2(2x^2-2x+1)-x(x-1)=1/2
2)=(2-1) (1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16) =(2^2-1)(2^2+1)...=1+2^32=4294967297