answer:Let’s go through the steps to demonstrate that ( fleft( frac{1}{19} right) + fleft( frac{1}{29} right) + cdots + fleft( frac{1}{n^2 + 7n + 11} right) > fleft( frac{1}{2} right) ). 1. From condition (2), we let ( y = -x ). Thus, we get: [ fleft( frac{1}{x} right) + fleft( -frac{1}{x} right) = f(0) = 0 ] This implies: [ fleft( -frac{1}{x} right) = -fleft( frac{1}{x} right) ] Hence, ( f(x) ) is an odd function. 2. Let ( -1 < x_1 < x_2 < 0 ). We find: [ f(x_1) - f(x_2) = f(x_1) + f(-x_2) = fleft( frac{frac{1}{x_1} - frac{1}{x_2}}{1 - frac{1}{x_1 x_2}} right) = fleft( frac{x_2 - x_1}{x_1 x_2 - 1} right) ] 3. Now note that: [ (1 + x_1)(1 - x_2) > 0 quad text{implies} quad x_1 - x_2 > x_1 x_2 - 1 quad text{so} quad -1 < frac{x_2 - x_2}{x_1 x_2 - 1} < 0 ] By (3), ( fleft( frac{x_2 - x_1}{x_1 x_2 - 1} right) > 0 ). Therefore, ( f(x_1) > f(x_2) ), so ( f(x) ) is decreasing on ( (-1,0) ). 4. Since ( f(x) ) is odd and positive on ( (-1,0) ), it is decreasing and negative on ( (0,1) ). 5. Consider the function ( fleft( frac{1}{k^2 + 7k + 11} right) ). We note: [ k^2 + 7k + 11 = (k+3)(k+4) - 1 ] Hence: [ frac{1}{k^2 + 7k + 11} = frac{1}{(k+3)(k+4) - 1} ] Applying the function: [ fleft( frac{1}{k^2 + 7k + 11} right) = fleft( frac{frac{1}{(k+3)(k+4)}}{1 - frac{1}{(k+3)(k+4)}} right) = fleft( frac{frac{1}{k+3} + left(-frac{1}{k+4}right)}{1 + frac{1}{k+3} cdot left(-frac{1}{k+4}right)} right) ] Using the property of ( f ): [ fleft( frac{frac{1}{k+3} + left(-frac{1}{k+4}right)}{1 + frac{1}{k+3} cdot left(-frac{1}{k+4}right)} right) = fleft( frac{1}{k+3} right) + fleft( -frac{1}{k+4} right) = fleft( frac{1}{k+3} right) - fleft( frac{1}{k+4} right) ] 6. Summing up these terms from ( k = 1 ) to ( n ): [ fleft( frac{1}{19} right) + fleft( frac{1}{29} right) + cdots + fleft( frac{1}{n^2 + 7n + 11} right) ] [ = left[ fleft( frac{1}{4} right) - fleft( frac{1}{5} right) right] + left[ fleft( frac{1}{5} right) - fleft( frac{1}{6} right) right] + cdots + left[ fleft( frac{1}{n+3} right) - fleft( frac{1}{n+4} right) right] ] [ = fleft( frac{1}{4} right) - fleft( frac{1}{n+4} right) ] 7. Since ( 0 < frac{1}{n+4} < 1 ) for ( n in mathbf{N}^+ ), and ( f(x) ) is negative on ( (0,1) ): [ fleft( frac{1}{n+4} right) < 0 implies fleft( frac{1}{4} right) - fleft( frac{1}{n+4} right) > fleft( frac{1}{4} right) > fleft( frac{1}{2} right) ] Therefore, the inequality holds: [ fleft( frac{1}{19} right) + fleft( frac{1}{29} right) + cdots + fleft( frac{1}{n^2 + 7n + 11} right) > fleft( frac{1}{2} right) ] [ boxed{} ]

answer:Given the equation a^{2}+2a-1=0, we are to find the value of the algebraic expression left(frac{{a^2}-1}{{a^2}-2a+1}-frac{1}{{1-a}}right) div frac{1}{{a^2}-a}. Let's break down the solution step by step: 1. **Simplify the given expression:** First, we recognize that the expression can be simplified by factoring and simplifying each component: [ left(frac{{a^2}-1}{{a^2}-2a+1}-frac{1}{{1-a}}right) div frac{1}{{a^2}-a} ] We know that a^2 - 1 = (a+1)(a-1) and a^2 - 2a + 1 = (a-1)^2. Also, a^2 - a = a(a-1). With these, we can rewrite the expression as: [ left[frac{(a+1)(a-1)}{(a-1)^{2}}+frac{1}{a-1}right] cdot a(a-1) ] This simplifies to: [ left(frac{a+1}{a-1}+frac{1}{a-1}right) cdot a(a-1) ] Further simplification gives: [ frac{a+1+1}{a-1} cdot a(a-1) ] Which simplifies to: [ a^{2}+2a ] 2. **Substitute the given equation:** Given that a^{2}+2a-1=0, we can solve for a^{2}+2a: [ a^{2}+2a = 1 ] 3. **Conclude the solution:** Therefore, substituting back into our original expression, we find that the value of the algebraic expression is: [ boxed{1} ] This step-by-step breakdown follows closely to the original solution, detailing each step of the simplification and substitution process, leading to the final answer.

answer:First, we use the formula for the area of a triangle, S_{triangle ABC}= frac {1}{2}bcsin A. Substituting the given values, we have sqrt {3}= frac {1}{2} times 1 times c times sin 60^{circ}, which simplifies to c=4. Next, we use the cosine rule, a^{2}=b^{2}+c^{2}-2bccdotcos A. Substituting the known values, we get a^{2}=1^{2}+4^{2}-2 times 1 times 4 times cos 60^{circ}=13, hence a= sqrt {13}. The inradius r of the triangle can be found using the formula r=frac{a}{2sin A}=frac{sqrt{13}}{2sin 60^{circ}}=frac{1}{3} sqrt {39}, hence 2r=frac{2}{3} sqrt {39}. Finally, we use the sine rule, frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=2R (where R is the circumradius), to get frac {a+b+c}{sin A+sin B+sin C}=frac{2rsin A+2rsin B+2rsin C}{sin A+sin B+sin C}=2r=boxed{frac{2}{3} sqrt {39}}.

answer:# Problem: [ text{在黑板上写有一个数: } 321321321321, text{擦去其中的某个或某些数字, 可以得到一个被} 9 text{整除的数, 这个数最大是} ] We'll go through the steps to determine the largest number divisible by 9 obtained by removing certain digits from 321321321321. 1. **Understanding the Requirement:** A number is divisible by 9 if the sum of its digits is divisible by 9. 2. **Calculate the Sum of All Digits:** The given number is 321321321321. [ text{Sum of the digits} = 3 + 2 + 1 + 3 + 2 + 1 + 3 + 2 + 1 + 3 + 2 + 1 ] Let's compute it step-by-step: [ 3 + 2 + 1 = 6 ] [ 3 + 2 + 1 = 6 ] [ 3 + 2 + 1 = 6 ] [ 3 + 2 + 1 = 6 ] Adding these results together: [ 6 + 6 + 6 + 6 = 24 ] 3. **Adjust the Sum to Make it Divisible by 9:** The sum of the digits 24 is not divisible by 9. We need to remove digits such that the remaining sum is divisible by 9. To achieve a sum divisible by 9, we can subtract either 3, 6, 12, 15, etc., from 24. The nearest is subtracting 6. Therefore, removing a combination of digits that sum to 6 will make the resulting number divisible by 9. The simplest way is to remove two digits: 3 and 3, or a single combination like 2+2+1+1. 4. **Removing Digits 3 + 3:** Let's remove each instance of '3' and check the resultant number. - Removing two '3's: [ 213212121 ] The sum of the digits: [ 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12 quad (text{which is divisible by 9}) ] 5. **Verifying Other Possibilities:** Finally, we have another possible combination like removing 2 and 1, but the considerations mention removing digits efficiently: - Removing one '1' and one '3': [ 32321211 ] While the sum doesn't aid exactly, In conclusion, - Properly the steps including remove more digits properly, [ remove 3+3, observes other may reannten exactly executable. Thus the largest possible combination: Conclusion [ boxed{3213212121} ]