answer:Since the function f(x)= begin{cases} 1-x，& xleq0 a^{x}，& x>0end{cases}, then f(-1)=2, f(1)=a, If f(1)=f(-1), then a=2, Therefore, the correct choice is boxed{text{B}}. From the piecewise function f(x), we can easily find the values of f(1) and f(-1), and then transform the equation f(1)=f(-1) into an equation about a. Combined with the range of values of the exponential function and the analytical expression of the piecewise function, solving the equation can yield the value of the real number a. This problem tests the knowledge of the values of piecewise functions and the comprehensive application of exponential functions. Among them, constructing an equation about a based on the properties of piecewise functions and exponential functions is the key to solving this problem.

answer:Given that the 5th term of an arithmetic sequence a_5=8, and a_1+a_2+a_3=6, We have the following system of equations: [ begin{cases} a_1 + 4d = 8 a_1 + a_1 + d + a_1 + 2d = 6 end{cases} ] Solving the system of equations, we get a_1=0 and d=2. Therefore, the answer is boxed{C}. By using the general term formula of an arithmetic sequence, we can set up a system of equations to find the common difference. This problem tests the basic knowledge of finding the common difference of an arithmetic sequence. When solving this problem, it is important to carefully read the question and make reasonable use of the properties of arithmetic sequences.

answer:To solve the equation, let's first move all terms to one side to get: x(5x+2) - 6(5x+2) = 0. Next, we factor by grouping: (x-6)(5x+2) = 0. From the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero. Thus, we have two possibilities: x - 6 = 0 quad text{or} quad 5x + 2 = 0. Solving each equation for x gives us the solutions: x = 6 quad text{and} quad x = -frac{2}{5}. Therefore, the solutions to the equation are x = boxed{6} and x = boxed{-frac{2}{5}}.

answer:A. |overrightarrow{a}|=|overrightarrow{b}| means the lengths of the vectors are equal, but the direction is not determined, so overrightarrow{a}=±overrightarrow{b} is not always true. B. If overrightarrow{a}//overrightarrow{b}, the directions of the two vectors are the same or opposite, but the lengths are not necessarily the same, so overrightarrow{a}=overrightarrow{b} is not always true. C. If overrightarrow{a}cdotoverrightarrow{b}=overrightarrow{b}cdotoverrightarrow{c}, it does not necessarily mean that overrightarrow{a}=overrightarrow{c}. D. If overrightarrow{a}//overrightarrow{b} and overrightarrow{b}//overrightarrow{c} (overrightarrow{b}neq 0), then it is true that overrightarrow{a}//overrightarrow{c}. Therefore, the correct answer is boxed{D}. This can be determined by understanding the relevant concepts of vectors and the properties of parallelism. This problem mainly tests the true or false judgment of propositions, involving the relevant concepts of vectors and the properties of vector parallelism, which is relatively basic.