answer:**Analysis** This question tests the application of abstract functions, function evaluation, and the application of basic inequalities. It is of medium difficulty. **Solution** Since the function f(x) satisfies f(x+1)= frac{1}{2}+ sqrt{f(x)-f^{2}(x)}, it follows that f(x) > 0 and f^{2}(x+1)= frac{1}{4}+ sqrt{f(x)-f^{2}(x)}+f(x)-f^{2}(x). Thus, f(x+1)-f^{2}(x+1)= frac{1}{2}+ sqrt{f(x)-f^{2}(x)}-left[frac{1}{4}+ sqrt{f(x)-f^{2}(x)}+f(x)-f^{2}(x)right]= frac{1}{4}-left[f(x)-f^{2}(x)right]. Therefore, f(x+1)-f^{2}(x+1)+f(x)-f^{2}(x)= frac{1}{4}. Let g(x)=f(x)-f^{2}(x), then g(x+1)+g(x)= frac{1}{4}. Hence, g(0)=g(2)=ldots=g(2016); g(1)=g(3)=ldots=g(2017); g(0)+g(2017)= frac{1}{4}. Therefore, f(0)-f^{2}(0)+f(2017)-f^{2}(2017)= frac{1}{4}. f(0)+f(2017)= frac{1}{4}+f^{2}(0)+f^{2}(2017) geqslant frac{1}{4}+ frac{[f(0)+f(2017)]^{2}}{2}. That is, 2[f(0)+f(2017)]^{2}-4[f(0)+f(2017)]+1leqslant 0. Solving this, we get f(0)+f(2017) in left[1-frac{sqrt{2}}{2}, 1+frac{sqrt{2}}{2}right]. Therefore, the correct choice is boxed{text{B}}.

answer:Let □ be x, △ be y, and ○ be z. From the problem, we have: - 2x + y + z = 17 (1) - x + 2y + z = 14 (2) - x + y + 2z = 13 (3) Subtracting equation (3) from equation (1), we get: x - y = 3 (4). Multiplying equation (2) by 2 and then subtracting equation (3), we get: x + 3y = 15 (5). Adding 3 times equation (4) to equation (5), we obtain: 4x = 24. Therefore, x = 6. Hence, the answer is boxed{6}.

answer:First, write the number in a form that suggests a binomial expansion: begin{align*} 256289062500 &= 2 cdot 50^8 + 8 cdot 50^7 + 28 cdot 50^6 + 56 cdot 50^5 &quad + 70 cdot 50^4 + 56 cdot 50^3 + 28 cdot 50^2 + 8 cdot 50 + 2. end{align*} Notice the coefficients are binomial coefficients. In fact, we have begin{align*} 256289062500 &= binom{8}{8} cdot 50^8 + binom{8}{7} cdot 50^7 + binom{8}{6} cdot 50^6 &quad + binom{8}{5} cdot 50^5 + binom{8}{4} cdot 50^4 + binom{8}{3} cdot 50^3 &quad + binom{8}{2} cdot 50^2 + binom{8}{1} cdot 50 + binom{8}{0}. end{align*} By the binomial theorem, this equals (50 + 2)^8 = 52^8. Therefore, the eighth root is boxed{52}.

answer:To solve the problem, we need to show that all polygons in the sequence have an area greater than ( frac{1}{3} ). Here's a detailed step-by-step proof: 1. **Initial Polygon**: - The initial polygon is a regular hexagon with area 1. 2. **Convexity**: - Each subsequent polygon remains convex. This is because each new polygon is formed by cutting off a corner and joining the adjacent edge midpoints, which continues to preserve convexity. 3. **Vertex Retention**: - It is crucial to note that each polygon in the sequence retains a vertex on each of the sides of the original hexagon. Hence, no side of the hexagon can be completely eliminated. 4. **Vertices Position**: - Let the original hexagon be labeled ( ABCDEF ). The midpoints of the sides are joined, and we consider the intersections of these segments to form a smaller hexagon. Specifically, the intersection points ( U, V, W, X, Y, Z ) form within the hexagon ( ABCDEF ). 5. **Intersection of Diagonals**: - The diagonals ( AC, BD, CE, DF, EA, FB ) from the original hexagon form the vertices of the inner hexagon ( UVWXYZ ). For example, point ( U ) is the intersection of diagonals ( AC ) and ( FB ). 6. **Segment Intersections**: - In any polygon from the sequence, vertices on segments like ( AB ) and ( BC ) must include the segment joining these two. This segment will always intersect portions of the diagonals such as ( BU ) and ( BV ), ensuring certain points on ( BU ) and ( BV ) are part of the polygon. 7. **Convex Hull**: - The points on segments ( BU, BV, CW, ) etc., guarantee that the convex hull of these points includes the inner hexagon ( UVWXYZ ). 8. **Area Calculation**: - We need to prove the area of ( UVWXYZ ) is ( frac{1}{3} ) of the area of ( ABCDEF ): - Recall that the triangle ( ABF ) in the hexagon is isosceles with an angle ( angle ABF = 30^circ ). - If ( AB = k ), then ( BF = ksqrt{3} ). - The segment ( BU ) equals ( frac{k}{sqrt{3}} ) because ( U ) is the midpoint of segment ( BF ). - Therefore, the distance ( UZ ) (a side of the inner hexagon) is given by ( ksqrt{3} - 2 cdot frac{k}{sqrt{3}} = frac{k}{sqrt{3}} ). 9. **Area Relation**: - Since each smaller hexagon maintains the ratio ( frac{1}{3} ) of the area of its predecessor due to the property of the side lengths and the constructed geometry, the area of hexagon ( UVWXYZ ) is ( frac{1}{3} ) the area of ( ABCDEF ). 10. **Conclusion**: - Given that the area of ( ABCDEF ) is 1, the area of ( UVWXYZ ) is ( frac{1}{3} ). We’ve demonstrated that every polygon derived in this sequence preserves an area at least ( frac{1}{3} ) of the original hexagon, hence achieving our objective. [ boxed{text{The area of any polygon in the sequence is greater than }} frac{1}{3}. ]