Question

急求2道数学题的答案过程

题目在这里，求过程！！！

Answer 1

1. (1) x^2-(x-1)=x^2-x+1=(x-1/2)^2+3/4>0恒成立
故x^2恒大于x-1, min=x-1
(2) ab/(4a^2+b^2)<=ab/2*根号下(4a^2*b^2)=ab/4ab=1/4
假设min>1/2, 则ab/(4a^2+b^2)>1/4, 与上式矛盾, 故min<=1/2

2. (1) S1-3=a1+a2+a3=a1*(1+q+q^2)
S4-6=a4+a5+a6=a1*(q^3+q^4+q^5)
S7-9=a7+a8+a9=a1*(q^6+q^7+q^8)
S7-9=q^3*S4-6 S4-6=q^3*S1-3
(2) S1-n, Sn+1-2n, S2n+1-3n...成等比数列
同上可得Sn+1-2n=q^n*S1-n, S2n+1-3n=q^n*Sn+1-2n

Answer 2

1.
(1)
证：
x^2-(x-1)
=x^2-x+1
=(x-1/2)^2+3/4>0
对任意实数x，恒有x^2>x-1
因此，当x∈R min{x^2，x-1}
(2)
证：
b/(4a^2+b^2)