很简单呢
首先按照他写的关系列式x(x+1)(x+2)[x(x+1)(x+2)+1]=3660
也就是说3660 可以分解成俩个自然数的乘积,且两个自然数相差1,即3660=60乘以61
所以X(x+1)(x+2)=60,即60可以分解成三个连续自然数,60=3乘4乘5 故x=3
(X△3)△2
=(X△3)×[(X△3)+1]
=x(x+1)(x+2)×[x(x+1)(x+2)+1]
=[x(x+1)(x+2)]^2+[x(x+1)(x+2)]
[x(x+1)(x+2)]^2+[x(x+1)(x+2)]-3660=0
[x(x+1)(x+2)-60][x(x+1)(x+2)-61]=0
x(x+1)(x+2)=60,x(x+1)(x+2)=61(舍去)
x(x+1)(x+2)=60=3×4×5
∴x=3
XΔ3=X(X+1)(X+2), (XΔ3)Δ2=[X(X+1)(X+2)][X(X+1)(X+2)+1]=3660=60×61, = ∴X(X+1)(X+2)=60=3×4×5. ∴X=3.