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a=根号17+1分之16=√17-1
(a+1)²=(√17)²
a²+2a-16=0
a5+2a4-17a3-a2+18a-17
=a^5+2a^4-16a^3-a^3-2a²+16a+a²+2a-16-1
=a³(a²+2a-16)-a(a²+2a-16)+(a²+2a-16)-1
=-1
a=(17-1)/(√17-1)=√17-1
a满足(a+1)²=17 所以a²+2a-16=0
a^5+2a^4-17*a³-a²+18a-17
=a³(a²+2a-16)-a³-a²+18a-17=-a³-a²+18a-17
=-a(a²+2a-16...
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a=(17-1)/(√17-1)=√17-1
a满足(a+1)²=17 所以a²+2a-16=0
a^5+2a^4-17*a³-a²+18a-17
=a³(a²+2a-16)-a³-a²+18a-17=-a³-a²+18a-17
=-a(a²+2a-16)+a²+2a-17=a²+2a-17
=a²+2a-16-1=0-1=-1
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回答第四题
设t=x^2,
so t^2-t-6=0
t=3, t=-2
t>=0,
t=3,x=+-根号3