初二数学题(急)
(x-1/4)(3x+3/4)(2x^2+1/8)
-(a-1/a)[(a+1/a)^2-1)](-a-1/a)[(a-1/a)^2+1]
参考答案:(x-1/4)(3x+3/4)(2x^2+1/8)
=6(x-1/4)(x+1/4)(x^2+1/16)
=6(x^2-1/16)(x^2+1/16)
=6(x^4-1/256)
=6x^4-3/128
-(a-1/a)[(a+1/a)^2-1)](-a-1/a)[(a-1/a)^2+1]
=-(a-1/a)[-(a+1/a)][a^2+1/a^2+2-1][a^2+1/a^2-2+1]
=(a^2-1/a^2)[(a^2+1/a^2)+1][(a^2+1/a^2)-1]
=(a^2-1/a^2)[(a^2+1/a^2)^2-1]
=(a^2-1/a^2)(a^4+1/a^4+1)
=a^6-1/a^6