answer:Solution: x^3 + 6x - x^2 - 6 = x^2(x - 1) + 6(x - 1) = (x^2 + 6)(x - 1), Since x > 1, then (x^2 + 6)(x - 1) > 0, Therefore, x^3 + 6x > x^2 + 6. So, the final answer is boxed{x^3 + 6x > x^2 + 6}.

answer:Given the equation |a^{3}-27|+left(b+2right)^{2}=0, we can break it down into two parts due to the properties of absolute values and squares:1. For the absolute value to result in 0, the expression inside must be 0. Therefore, we have:[|a^{3}-27|=0][a^{3}-27=0]Solving for a gives us:[a^{3}=27][a=3]2. For the square to result in 0, the expression inside the square must also be 0. Thus, we have:[left(b+2right)^{2}=0][b+2=0]Solving for b gives us:[b=-2]Adding the values of a and b together:[a+b=3+(-2)][a+b=1]Therefore, the solution to the problem is:[boxed{1}]

answer:[Solution] After calculation, the first few terms of the sequence left{x_{n}right} area, b, b-a,-a,-b, a-b, a, b, b-a, -a, quad-b, quad a-b, cdotsFrom this, we can see that x_{n+6}=x_{n}, meaning left{x_{n}right} is a sequence with a period of 6.begin{aligned}therefore & x_{100}=x_{6 times 16+4}=x_{4}=-a . text { Also, } & x_{6 k+1}+x_{6 k+2}+x_{6 k+3}+x_{6 k+4}+x_{6 k+5}+x_{6 k+6} = & x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} = & a+b+(b-a)+(-a)+(-b)+(a-b)=0, therefore quad & S_{100}=S_{6 times 16+4}=S_{4}=a+b+(b-a)+(-a) & =2 b-a .end{aligned}Therefore, the answer is (A).

answer:Since the set A={xinmathbb{R}|x^{2}-3xleqslant 0}={x|0leqslant xleqslant 3}, and B={1,2}, therefore, Acap B={1,2}. Hence, the correct choice is: boxed{B}. First, we find the sets A and B separately, and then use the definition of intersection to find Acap B. This question tests the method of finding intersections, which is a basic problem. When solving it, one should carefully read the question and apply the definition of intersection appropriately.

answer:[Solution] According to the condition 1000 leqslant n leqslant 2000, we geta_{n} geqslant sqrt{57121+35 times 1000}=303.51441 cdotsand a_{n} leqslant sqrt{57121+35 times 2000}=356.54032 cdots,which means quad 304 leqslant a_{n} leqslant 357Also, since 57121=1632 times 35+1,we have a_{n}=sqrt{35(n+1632)+1}.Thus, a_{n}^{2}-1=left(a_{n}-1right)left(a_{n}+1right) can be divided by 35=5 times 7.Therefore, 7 mid left(a_{n}-1right) or 7 mid left(a_{n}+1right).That is, a_{n}=7 k+1 or a_{n}=7 k-1 .(k in mathbb{Z})In the first case, 304 leqslant 7 k+1 leqslant 357, which means 44 leqslant k leqslant 50. Using a calculator to compute a_{n}^{2}-1=(7 k+1)^{2}-1 for k=44,45, cdots, 50, and selecting those divisible by 5, we get four values of n=left[frac{a_{n}^{2}-1}{35}right]-1632: 1096left(k=44, a_{n}=309right), 1221 left(k=45, a_{n}=316right), 1749 left(k=49, a_{n}=344right), and 1888(k=50, left.a_{n}=351right).In the second case, a_{n}=7 k-1 and 44 leqslant k leqslant 51. Similarly, we get four values of n: 1185left(k=45, a_{n}=314right), 1312 left(k=46, a_{n}=321right), 1848 left(k=50, a_{n}=349right), and 1989left(k=51, a_{n}=356right).

answer:In every triangle with sides a, b, c and inradius r, it holds (as one can see from a sketch)frac{1}{2} r(a+b+c)=A quad text { thus } quad r=frac{2 A}{a+b+c}(here A denotes the area of the triangle). If the triangle is right-angled and c is the hypotenuse, thena b=2 A quad text { thus } quad r=frac{a b}{a+b+c} quad text { and } quad R=frac{c}{2}(by the Thales' theorem). Therefore,frac{R}{r}=frac{c(a+b+c)}{2 a b}=frac{1}{2}left[frac{c}{sqrt{a b}} cdot frac{a+b}{sqrt{a b}}+frac{c^{2}}{a b}right]=frac{1}{2}left[sqrt{frac{a^{2}+b^{2}}{a b}}+frac{a+b}{sqrt{a b}}+frac{a^{2}+b^{2}}{a b}right]It always holds that (a-b)^{2} geq 0, thus a^{2}+b^{2} geq 2 a b and for a b>0 : frac{a^{2}+b^{2}}{a b} geq 2, similarly frac{a+b}{sqrt{a b}} geq 2. From this it followsfrac{R}{r} geq frac{1}{2}(2 sqrt{2}+2)=sqrt{2}+1