Question

几道初一数学题~

1、已知（2a-1)^2+绝对值3b-1绝对值=0，求5(3a^2-ab^2)-(ab^2+3a^2*b)的值
2、已知a^2-3a+1=0,求a+1/a的值
3、(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1(n是正整数）

Answer 1

1、已知（2a-1)^2+绝对值3b-1绝对值=0，求5(3a^2-ab^2)-(ab^2+3a^2*b)的值

（2a-1)^2+∣3b-1∣=0
a=1/2
b=1/3

5(3a^2-ab^2)-(ab^2+3a^2*b)
=5a(3a-b^2)-ab(b+3a)
=5*1/2(3*1/2-1/9)-1/2*1/3(1/3+3*1/2)
=5/2(3/2-1/9)-1/6(1/3+3/2)
=5/2(27/18-2/18)-1/6(2/6+9/6)
=5/2*25/18-1/6*11/6
=125/36-11/36
=114/36
=38/12
=19/6

2、已知a^2-3a+1=0,求a+1/a的值

a^2-3a+1=0(等式两边同时除以a)
a-3+1/a=0
a+1/a=3

3、(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1(n是正整数

(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2^2-1)(2^2+1)(2^4+1)…(2^2n+1)+1
=(2^4-1)(2^4+1)…(2^2n+1)+1
=(2^8-1)…(2^2n+1)+1
.....
=(2^2n-1)(2^2n+1)+1
=2^4n-1+1
=2^4n

Answer 2

第二题等于5 第一和三题目有说清楚吗