Question

已知满足(-8的a次方)的2次方÷4的5-2b=2的1-b的2次方，9的a次方的b次方=3的b次方的a次方÷27

（1）求a的2次方+b的2次方的值
（2）计算[ab/(a+b-1)(a+b)] +[（ab/（a+b)(a+b+1)]+[（ab/（a+b+1)(a+b+2)]+...+[(ab/(a+b+99)（a+b+100）]

Answer 1

1) (-8^a)^2÷4^(5-2b)=[2^(1-b)]^2
2^(6a)=2^(2-2b)*2^(10-4b)
2^(6a)=2^(12-6b)
6a=12-6b
a+b=2 ①
(9^a)^b=(3^b)^a÷27
3^(2ab)=3^(ab-3)
2ab=ab-3
ab=-3 ②
a^2+b^2=(a+b)^2-2ab=10

2) [ab/(a+b-1)(a+b)] +[ab/(a+b)(a+b+1)]+[ab/(a+b+1)(a+b+2)]+...+[(ab/(a+b+99)(a+b+100)]

=ab[1/(a+b-1)(a+b)] +[1/(a+b)(a+b+1)]+[1/(a+b+1)(a+b+2)]+...+[(1/(a+b+99)(a+b+100)]
=ab[1/(a+b-1)-1/(a+b)+1/(a+b)-1/(a+b+1)+1/(a+b+1)-1/(a+b+2)+...+1/(a+b+99)-1/(a+b+100)]
=ab[1/(a+b-1)-1/(a+b+100)]
=101ab/[(a+b-1)(a+b+100)]
=101*(-3)/[(2-1)*(2+100)]
=-101/34