answer:To solve the problem, we visualize triangle ABC with angle A = 50^circ, angle B = 70^circ, angle C = 60^circ, and triangle XYZ inside, with Gamma as both the incircle of triangle ABC and circumcircle of triangle XYZ. For triangle ABC, since Gamma is the incircle, it implies that each segment from a vertex to the point where the incircle touches is equal. For instance, if the incircle touches overline{BC} at X, overline{AB} at Y, and overline{AC} at Z, then overline{AX}, overline{BY}, and overline{CZ} are each split into two equal segments due to the properties of the angle bisectors and the incircle. This property establishes that triangle XYZ is likely equilateral because each segment to the circle's touch is equal; thus, each angle of triangle XYZ is 60 degrees due to the properties of an equilateral triangle embedded in another triangle respecting angle bisectors. The calculation for angle YZX can be deduced from the symmetry and equal split mentioned: 1. Gamma touches overline{BC}, overline{AB}, and overline{AC} equally. 2. Each segment's symmetry means triangle XYZ is equilateral. 3. Therefore, angle YZX = 60^circ. This is succinctly boxed as boxed{60^circ}.

answer:**Analysis** According to the problem, the correlation coefficient r_{1} is negatively correlated, and the correlation coefficient r_{2} is positively correlated, from which the conclusion is drawn. This question tests the application of judging the linear correlation coefficient, which is a basic problem. **Solution** Based on the problem, the starting smoking age (X) and the relative risk of developing lung cancer (Y) have a negative correlation, so the correlation coefficient r_{1} < 0; The daily smoking (U) cigarettes and the relative risk of developing lung cancer (V) have a positive correlation, so the correlation coefficient r_{2} > 0; Therefore, r_{1} < 0 < r_{2}. Hence, the correct choice is: boxed{A}.

answer:1. **Expression Simplification**: Begin by noting that (sqrt{x^2 + 1}^2 = x^2 + 1), so the expression simplifies to: [ x^2 + (x^2 + 1) - frac{1}{x^2 + (x^2 + 1)} ] Simplifying further: [ 2x^2 + 1 - frac{1}{2x^2 + 1} ] 2. **Simplify the Fraction**: Rationalize the denominator of the subtraction term: [ frac{1}{2x^2 + 1} = frac{1}{2 (x^2 + tfrac{1}{2})} ] Let's substitute this back into the main expression: [ (2x^2 + 1) - frac{1}{2x^2 + 1} ] 3. **Evaluation of Rationality**: The term (2x^2 + 1) is rational as (x^2) is always rational given (x) is rational; however, the fraction (frac{1}{2x^2 + 1}) introduces complexity. For this to be rational, (2x^2 + 1) must not only be rational but specifically an integer that does not zero the denominator. Thus, (x) must be rational for the entire expression to remain rational. 4. **Conclusion**: (2x^2 + 1 - frac{1}{2x^2 + 1} = k) where (k) is rational, implying (x) must be rational and (2x^2 + 1) must not disrupt the denominator. [text{The set of all rational numbers x}] The final answer is boxed{B (rational (x))}

answer:First, let's translate the problem and the solution from Chinese to English. The problem asks us to find the sum of the cubes of the first six positive integers, given a pattern in the sums of cubes of the first few positive integers. The provided solution states that according to the problem, (1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+...+n^{3}=(1+2+...+n)^{2}), so (1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}=21^{2}). Now, let's format the solution using LaTeX and present it in a step-by-step manner. 1. Observe the given pattern: 1^{3}+2^{3}=3^{2} 1^{3}+2^{3}+3^{3}=6^{2} 1^{3}+2^{3}+3^{3}+4^{3}=10^{2} 2. From the pattern, we can infer the general rule: 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+...+n^{3}=(1+2+...+n)^{2} 3. Apply the general rule to find the sum of the cubes of the first six positive integers: 1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}=(1+2+3+4+5+6)^{2} 4. Calculate the sum in the parentheses: 1+2+3+4+5+6=21 5. Square the result: 21^{2}=441 So, the sum of the cubes of the first six positive integers is boxed{441}.