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解:a²-3a+1=0.------------(1)
b²-3b+1=0.------------(2)
(1)-(2),得:a²-b²-3a+3b=0, (a-b)(a+b)-3(a-b)=0, (a-b)(a+b-3)=0.
又a≠b,则:a-b≠0, a+b-3=0, 得:a+b=3;
(1)+(2),得:a²+b²-3a-3b+2=0, a²+b²-3(a+b)+2=0, a²+b²-3x3+2=0, a²+b²=7.
则:(a+b)²-2ab=7, 3²-2ab=7, 得:ab=1.
∴1/(1+b²)+1/(1+a²)
=(1+a²)/[(1+a²)(1+b²)]+(1+b²)/[(1+a²)(1+b²)]
=(2+a²+b²)/[1+a²+b²+(ab)²]
=(2+7)/(1+7+1²)
=1
1年前
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