- 初一数学题求6(7+1)(7平方+1)(7四次方+1)(7八次方
- 求 6(7+1)(7平方+1)(7四次方+1)(7八次方+1)+1的值
已知:a+1/a=2,求a平方＋１／a平方的值
- 1. 6*(7+1)*(7^2+1)*(7^4+1)*(7^8+1)+1
=(7-1)*(7+1)*(7^2+1)*(7^4+1)*(7^8+1)+1
=(7^2-1)*(7^2+1)* (7^4+1)*(7^8+1)+1
=(7^4-1)*(7^4+1)* (7^8+1)+1
=(7^8-1)*(7^8+1)+1
=7^16
2.由已知求出a=1
原式=2