Question

求助数学大神，这道题进一步追问的两问怎么做？书上没给答案，作业帮也搜不到，求大神帮忙，谢谢～

Answer 1

(2)设a=n-1,b=n,c=n+1,C=3A,

cosA=(b^2+c^2-a^2)/(2bc)=（n+4）/[2(n+1)],
cosC=(a^2+b^2-c^2)/(2ab)=(n-4)/[2(n-1)],
cosC=4(cosA)^3-3cosA,
所以(n-4)/[2(n-1)]=4{（n+4）/[2(n+1)]}^3-3（n+4）/[2(n+1)],
两边都乘以2(n-1)(n+1)^3,得
(n-4)(n+1)^3=(n-1)(n+4)^3-3(n+4)(n-1)(n+1)^2,
(n^2-3n-4)(n^2+2n+1)=(n^2+3n-4)(n^2+8n+16)-3(n^2-1)(n^2+5n+4),
n^2(n^2+2n+1-n^2-8n-16+3n^2+15n+12)-3n(n^2+2n+1+n^2+8n+16)
+4(n^2+8n+16-n^2-2n-1)-3n^2-15n-12=0,
n^2(3n^2+9n-3)-3n(2n^2+10n+17)+4(6n+15)-3n^2-15n-12=0,
n^2(n^2+3n-1)-n(2n^2+10n+17)+4(2n+5)-n^2-5n-4=0,
整理得f(n)=n^4+n^3-12n^2-14n+16=0,

f(4)=88,f(3)=-26,f(2)=-36,f(1)=-8,f(0)=16,
所以f(n)无正整数零点，即此类三角形不存在。