关于乘法公式的算式题,要正确的啊（1-X/2)(1+x/2)(1+x^2/4)(1+x^4/16)（y-2x)(-2x-y)998*1002-999*1001100^2-99^2+98^2-97^2...+2^2-1^2(1-1/2^2)(1-1/3^2)(1-1/4^2)...(1-1/10^2)(3+1)(3^2+1)(3^4+1)(3^8+1)-3^16/2

（1-x/2)(1+x/2)(1+x^2/4)(1+x^4/16)
=（1 - x^2/4)(1 + x^2/4)(1 + x^4/16)
=(1 - x^4/16))(1+x^4/16)
=1 - x^8/256
（y-2x)(-2x-y)
=-（y-2x)(y+2x)
=-(y^2 - 4x^2)
=4x^2 - y^2
998*1002-999*1001
=(1000-2)(1000+2)-(1000-1)(1000+1)
=[(1000)^2 - 4]-[(1000)^2 - 1]
=-3
100^2-99^2+98^2-97^2...+2^2-1^2
=(100+99)(100-99)+(98+97)(98-97)+.....+(2+1)(2-1)
=(100+99)+(98+97)+.....+1
=5050
(1-1/2^2)(1-1/3^2)(1-1/4^2)...(1-1/10^2)
=(1+ 1/2)(1- 1/2)(1+ 1/3)(1- 1/3)(1+ 1/4)(1- 1/4).....(1+ 1/10)(1 -1/10)
=(1/2)*(11/10)
=11/20
(3+1)(3^2+1)(3^4+1)(3^8+1)-3^16/2
=[(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)/(3-1)]- 3^16/2
=[(3^2-1)(3^2+1)(3^4+1)(3^8+1)/2] - 3^16/2
=[(3^4-1)(3^4+1)(3^8+1)/2] - 3^16/2
=[(3^8-1)(3^8+1)/2]- 3^16/2
=[(3^16 - 1)/2]- 3^16/2
=(3^16/2) -1/2 - 3^16/2
=-1/2

(1-X/2)(1+x/2)(1+x^2/4)(1+x^4/16)
=[1-(X^2)/4][1+(X^2)/4]
=[1-(X^4)/16]
(y-2x)(-2x-y)
=-(y-2x)(y+2x)
=-(y^2-4x^2)
=4x^2-y^2
998×1002-999×1001
=(1000-2)×(1000+2)-(1000-1)×(1000+1)
=1000000-4-1000000+1
=-3
100^2-99^2+98^2-97^2……+2^2-1^2
=(100+99)(100-99)+(98+97)(98-97)+……(2+1)(2-1)
=199+195+……+3
=(3+199)×99÷2
=202×99÷2
=101×99
=9999

=（1-x^2/4)(1+x^2/4)(1+x^4/16)==（1-x^4/16)(1+x^4/16)=1-x^8/256=-（y-2x)(2x+y)=(2x-y)(2x+y)=4x^2-y^2=1000*1002-2*1002-1000*1001+1001=1000*(1002-1001)+1001-2*1002=1000+1001-2004=-3=(100-99)(100+99)+(98...