UC知道规律题目``咋做`?
来源：百度知道 编辑：UC知道 时间：2024/09/24 04:37:05
①(x-y)(x+y)=_____
②(x-y)(x+y)(x^2+y^2)=________
④(x-y)(x+y))(x^2+y^2))(x^4+y^4)=______
像这样下去,第五个式子是什么?并计算8(3^2+1)(3^4+1)…(3^32+1)的值.
②(x-y)(x+y)(x^2+y^2)=________
④(x-y)(x+y))(x^2+y^2))(x^4+y^4)=______
像这样下去,第五个式子是什么?并计算8(3^2+1)(3^4+1)…(3^32+1)的值.
①(x-y)(x+y)=x^2-y^2
②(x-y)(x+y)(x^2+y^2)=x^4-y^4
④(x-y)(x+y))(x^2+y^2))(x^4+y^4)=x^8-y^8
第五个式子是(x-y)(x+y))(x^2+y^2))(x^4+y^4)(x^8+y^8)(x^16+y^16)=(x^32-y^32)
8(3^2+1)(3^4+1)…(3^32+1)
=[8(3^2-1)(3^2+1)(3^4+1)…(3^32+1)]/(3^2-1)
=8(3^64-1)/(3^2-1)
=3^64-1
根据平方差公式啊!
①(x-y)(x+y)=x^2-y^2
②(x-y)(x+y)(x^2+y^2)=x^4-y^4
④(x-y)(x+y))(x^2+y^2))(x^4+y^4)=x^8-y^8
第五个式子:
8(3^2+1)(3^4+1)…(3^32+1)
=8[(3^2-1)(3^2+1)(3^4+1)…(3^32+1)]/(3^2-1)
_______ _____
=8(3^64-1)/(3^2-1)
=3^64-1