问道数学题!~高手帮帮我

2025-04-19 00:37:07
推荐回答(3个)
回答1:

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K
2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K
2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

回答2:

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K

参考证明:http://zhidao.baidu.com/question/558849.html

回答3:

2^(6n-3) + 3^(2n-1)
= 2^[3*(2n-1)] + 3^(2n-1)
= (2^3)^(2n-1) + 3^(2n-1)
= 8^(2n-1) + 3^(2n-1)

x = 8
y = 3

x^(2n-1) + y(2n-1)
= (x + y)[x^(2n-2) - x^(2n-3)*y + .... - y^(2n-2)]
= (x + y)*K
= 11K