answer:We start by assuming ax^2+18x+9 is the square of a binomial, which can be written as (rx+s)^2. Expanding this, we have: [ (rx+s)^2 = r^2x^2 + 2rsx + s^2. ] For this expression to match ax^2+18x+9, we need: - The coefficient of x^2 in both expressions to be equal, so a = r^2. - The coefficient of x to match, so 2rs = 18. - The constant terms to be equal, so s^2 = 9. From s^2 = 9, we have s = 3 or s = -3. We choose s = 3 for simplicity; the process for s = -3 will be analogous. Now, substituting s = 3 into 2rs = 18 gives: [ 2r cdot 3 = 18 implies 6r = 18 implies r = 3. ] Thus, our binomial becomes (3x + 3)^2, which expands to: [ (3x+3)^2 = 9x^2 + 18x + 9. ] Comparing with the original expression ax^2+18x+9, we find a = 9. Therefore, a = boxed{9}.

answer:To find the greatest positive integer ( k ) such that the inequality holds for all ( a, b, c in mathbb{R}^+ ) satisfying ( abc = 1 ): [ frac{1}{a} + frac{1}{b} + frac{1}{c} + frac{k}{a+b+c+1} geq 3 + frac{k}{4} ] we start by substituting ( a = t, b = t, c = frac{1}{t^2} ) for ( t neq 1 ). This substitution maintains the condition ( abc = 1 ). 1. Substitute ( a = t, b = t, c = frac{1}{t^2} ) into the inequality: [ frac{1}{t} + frac{1}{t} + t^2 + frac{k}{2t + frac{1}{t^2} + 1} geq 3 + frac{k}{4} ] 2. Simplify the left-hand side: [ 2 cdot frac{1}{t} + t^2 + frac{k}{2t + frac{1}{t^2} + 1} = frac{2}{t} + t^2 + frac{k}{2t + frac{1}{t^2} + 1} ] 3. Rearrange the inequality: [ frac{2}{t} + t^2 - 3 geq k left( frac{1}{4} - frac{t^2}{2t + frac{1}{t^2} + 1} right) ] 4. Simplify the right-hand side: [ frac{2}{t} + t^2 - 3 geq k left( frac{1}{4} - frac{t^2}{2t^3 + t^2 + 1} right) ] 5. Multiply both sides by ( 4(2t^3 + t^2 + 1) ): [ 4(2t^3 + t^2 + 1) left( frac{2}{t} + t^2 - 3 right) geq k left( 4(2t^3 + t^2 + 1) left( frac{1}{4} - frac{t^2}{2t^3 + t^2 + 1} right) right) ] 6. Simplify further: [ 4(2t^3 + t^2 + 1) left( frac{2}{t} + t^2 - 3 right) geq k left( (2t^3 + t^2 + 1) - 4t^2 right) ] 7. Choose ( t = frac{2}{3} ): [ 4 left( 2 left( frac{2}{3} right)^3 + left( frac{2}{3} right)^2 + 1 right) left( frac{2}{frac{2}{3}} + left( frac{2}{3} right)^2 - 3 right) geq k left( 2 left( frac{2}{3} right)^3 + left( frac{2}{3} right)^2 + 1 - 4 left( frac{2}{3} right)^2 right) ] 8. Simplify the expression: [ 4 left( frac{16}{27} + frac{4}{9} + 1 right) left( 3 + frac{4}{9} - 3 right) geq k left( frac{16}{27} + frac{4}{9} + 1 - frac{16}{9} right) ] 9. Calculate the values: [ 4 left( frac{16}{27} + frac{12}{27} + frac{27}{27} right) left( frac{4}{9} right) geq k left( frac{16}{27} + frac{12}{27} + frac{27}{27} - frac{48}{27} right) ] [ 4 left( frac{55}{27} right) left( frac{4}{9} right) geq k left( frac{7}{27} right) ] 10. Simplify further: [ frac{880}{63} geq k ] Since ( k ) must be an integer, the largest possible value for ( k ) is 13. We need to verify that ( k = 13 ) satisfies the original inequality. 11. Verify ( k = 13 ): [ frac{1}{a} + frac{1}{b} + frac{1}{c} + frac{13}{a+b+c+1} geq 3 + frac{13}{4} ] 12. Substitute ( a = t, b = t, c = frac{1}{t^2} ): [ frac{2}{t} + t^2 + frac{13}{2t + frac{1}{t^2} + 1} geq 3 + frac{13}{4} ] 13. Simplify and verify: [ frac{2}{t} + t^2 + frac{13}{2t + frac{1}{t^2} + 1} geq 6.25 ] By verifying the inequality for ( k = 13 ), we conclude that the greatest positive integer ( k ) is 13. The final answer is ( boxed{13} )

answer:1. Since f(x) = 2sqrt{3}sin xcos x + cos 2x + 3 = sqrt{3}sin 2x + cos 2x + 3 = 2sinleft(2x + frac{pi}{6}right) + 3, the smallest positive period of the function f(x) is frac{2pi}{2} = pi. Therefore, the smallest positive period of f(x) is boxed{pi}. 2. Since x in left[0, frac{pi}{4}right], we have 2x + frac{pi}{6} in left[frac{pi}{6}, frac{2pi}{3}right]. Therefore, when 2x + frac{pi}{6} = frac{pi}{2}, the function f(x) attains its maximum value of 5. At this time, x = frac{pi}{6}. Thus, the maximum value of the function f(x) in the interval left[0, frac{pi}{4}right] is boxed{5}, and the corresponding value of x is boxed{frac{pi}{6}}.

answer:1. **Define the variables and the setup:** - Let ( S ) denote the area of triangle ( AOD ). - Define ( x = AO ) and ( y = DO ). - Let ( a = AB ), ( b = BC ), ( c = CD ), and ( d = DA ). - Let ( k ) be the coefficient of similarity between triangles ( BOC ) and ( AOD ). 2. **Calculate the sums involving the radii:** The formula for the sum of the reciprocals of the inradii for triangle areas and semiperimeters is given by: [ 2 left( frac{1}{r_1} + frac{1}{r_3} right) = frac{d + x + y}{S} + frac{k d + k x + k y}{k^2 S} ] Similarly, we can express the sum involving ( r_2 ) and ( r_4 ): [ 2 left( frac{1}{r_2} + frac{1}{r_4} right) = frac{a + x + k y}{k S} + frac{c + k x + y}{k S} ] 3. **Relate the areas of triangles ( BOC ) and ( AOD ), as well as areas of triangles ( AOB ) and ( COD ):** - Since ( S_{BOC} = k^2 S ) and ( S_{AOB} = S_{COD} = k S ), we obtain: [ frac{x + y}{S} + frac{x + y}{k^2 S} = frac{x + k y}{k S} + frac{k x + y}{k S} ] 4. **Verify the equality:** Consider: [ frac{d + x + y}{S} + frac{k d + k x + k y}{k^2 S} ] and [ frac{a + x + k y}{k S} + frac{c + k x + y}{k S} ] By simplifying the terms, we eventually reach: [ frac{d + x + y}{S} + frac{k d + k x + k y}{k^2 S} = frac{a + x + k y}{k S} + frac{c + k x + y}{k S} ] 5. **Concluding step:** Hence, we establish that: [ frac{1}{r_1} + frac{1}{r_3} = frac{1}{r_2} + frac{1}{r_4} ] which completes the proof. (blacksquare)