answer:To find the probability that both drawn balls are black, we can use the formula for conditional probability. The probability of drawing a black ball on the first draw is the number of black balls divided by the total number of balls. After drawing one black ball, the probability of drawing a second black ball is the number of remaining black balls divided by the total number of remaining balls. Let's calculate it step by step: 1. Probability of drawing the first black ball (P(B1)): [ P(B1) = frac{text{Number of black balls}}{text{Total number of balls}} = frac{10}{10 + 5} = frac{10}{15} = frac{2}{3} ] 2. Probability of drawing the second black ball after the first one has been drawn (P(B2|B1)): [ P(B2|B1) = frac{text{Number of remaining black balls}}{text{Total number of remaining balls}} = frac{9}{14} ] Now, to find the probability of both events happening (drawing two black balls in a row), we multiply the probabilities of each event: [ P(text{Both black}) = P(B1) times P(B2|B1) ] [ P(text{Both black}) = frac{2}{3} times frac{9}{14} ] [ P(text{Both black}) = frac{2 times 9}{3 times 14} ] [ P(text{Both black}) = frac{18}{42} ] [ P(text{Both black}) = frac{3}{7} ] So, the probability that both drawn balls are black is boxed{(frac{3}{7})} .

answer:Part 1: Solve the inequality ( x^{log_a x} > frac{x^{frac{9}{2}}}{a^2} ) 1. Start by taking the logarithm base (a) of both sides of the inequality: [ log_aleft( x^{log_a x} right) > log_aleft( frac{x^{frac{9}{2}}}{a^2} right) ] 2. Use the properties of logarithms. Specifically, (log_a(x^b) = b log_a(x)) and (log_a(frac{b}{c}) = log_a(b) - log_a(c)): [ log_a x cdot log_a x > log_aleft( x^{frac{9}{2}} right) - log_a(a^2) ] 3. Simplifying log expressions: [ (log_a x)^2 > frac{9}{2} log_a x - 2 ] 4. Rewrite the inequality in standard quadratic form: [ (log_a x)^2 - frac{9}{2} log_a x + 2 > 0 ] 5. Factor the quadratic expression: [ left(log_a x - 4right)left(log_a x - frac{1}{2}right) > 0 ] 6. Evaluate the inequality by considering the sign of the factors. The product is positive if both factors are positive or both factors are negative. - Case 1: (log_a x - 4 > 0) and (log_a x - frac{1}{2} > 0) [ log_a x > 4 ] - Case 2: (log_a x - 4 < 0) and (log_a x - frac{1}{2} < 0) [ log_a x < frac{1}{2} ] 7. Combine the results: [ log_a x > 4 quad text{or} quad log_a x < frac{1}{2} ] 8. Translate back to x: - For (a > 1), (x > a^4) or (0 < x < a^{frac{1}{2}}). [ boxed{x > a^4 , text{or} , 0 < x < a^{frac{1}{2}}} ] - For (0 < a < 1), (log_a x < 4 quad text{and} quad log_a x > frac{1}{2}): [ frac{1}{2} < log_a x < 4 ] 9. Translate to x for (0 < a < 1): [ a^4 < x < a^{frac{1}{2}} ] Part 2: Solve the equation ((sqrt{5+2sqrt{6}})^{x}+(sqrt{5-2sqrt{6}})^{x}=10) 1. Let ( (sqrt{3} + sqrt{2})^x = A) and ((sqrt{3} - sqrt{2})^{ - x } = B ). Notice symmetry and properties to reduce complexity. 2. Multiply both sides of the equation by ((sqrt{3} + sqrt{2})^x): [ (sqrt{3} + sqrt{2})^{2x} + (sqrt{3} - sqrt{2})^x(sqrt{3} + sqrt{2})^x = 10(sqrt{3} + sqrt{2})^x ] 3. Notice the combination of terms based on properties: [ (sqrt{3} + sqrt{2})^{2x} + 1 = 10(sqrt{3} + sqrt{2})^x ] 4. Simplify to a quadratic in terms of ((sqrt{3} + sqrt{2})^x), let (y = (sqrt{3} + sqrt{2})^x), transforming the equation to: [ y^2 - 10y + 1 = 0 ] 5. Solve this quadratic equation: [ y = frac{10 pm 2sqrt{24}}{2} ] Simplify the solution to exact form: [ y = 5 pm 2sqrt{6} ] 6. Then, ( (sqrt{3} + sqrt{2})^x = 5 + 2sqrt{6} ) or ( 5 - 2sqrt{6} ): - If ( sqrt{3}+sqrt{2}=5+2sqrt{6}quad x=2 ) - If ( sqrt{3}-sqrt{2}=}"); 7. Recognize symmetry, setting ( x=-2 ) in finite cases: [ eq:i=2 quad text{or} quad -2^{-2} ] # Thus, the solutions are: ( boxed{x = 2 , text{or} , x = -2} )

answer:Given the problem, we have to find the angles and sides in a tetrahedron and prove a specific statement involving the triangles and their angles. We will work step-by-step to reach the conclusion: Part 1: Determine the Intersection Angle 1. **Identify Points and Planes**: - We know that E is a point on the edge AA_1, and A E = frac{3-sqrt{6}}{3}. - The plane alpha passes through points E, C, and D_1. 2. **Common Points and Lines**: - Plane alpha intersects plane D C C_{1} D_{1} in line CD_{1} as both planes share this line. - Also, plane alpha intersects plane A B B_{1} A_{1} at point E. 3. **Parallel Lines and Trapezoid**: - Since E F is parallel to CD, we form a trapezoid E F C D_{1}. 4. **Perpendicular and Lengths**: - Draw EH parallel to AD, intersecting DD_{1} at H such that EH perp to plane D C C_{1} D_{1}. - Given EH = 1, we next find the segment HP in the plane D C C_{1} D_{1} perpendicular to CD_1 at P. 5. **Using Coordinate Geometry**: - Using A_1 E = 1 - AE = frac{sqrt{6}}{3}, - We find H P = frac{1}{sqrt{2}}D_{1} H = frac{sqrt{3}}{3} from geometric relations. 6. **Angle Calculation**: - Using tan angle E P H = frac{E H}{H P}, we get tan angle E P H = sqrt{3}. - Hence, angle E P H = 60^circ. Part 2: Prove the Given Relation 7. **Triangle A B C Details**: - Let the edges of triangle A B C be a, b, and c. 8. **Using Sine Rule**: - In triangle A B F, we use the sine rule, [ A E = frac{c sin beta}{sin (alpha + beta)} ] 9. **Calculation using Trigonometric Identities**: - Using A E = frac{2R sin beta sin 3gamma}{sin left( frac{pi}{3} - gamma right)} - and the triple angle identity for sine: [ sin 3gamma = 3 sin gamma - 4 sin^3 gamma = 4 sin gamma left( frac{3}{4} - sin^2 gamma right) = 4 sin gamma left( sin^2 frac{pi}{3} - sin^2 gamma right) ] - Hence, [ sin 3gamma = 2 sin gamma (cos 2gamma - cos frac{2pi}{3}) = 4 sin gamma sin left( frac{pi}{3} + gamma right) sin left( frac{pi}{3} - gamma right) ] 10. **Determining AF**: - So, [ AF = 8 R sin beta sin gamma sin left( frac{pi}{3} + gamma right) ] 11. **Same Calculation for AE**: - Similarly, [ AE = 8 R sin beta sin gamma sin (frac{pi}{3} + beta) ] 12. **Using Law of Cosines in triangle A E F**: - We apply the cosine rule: [ E F^2 = A E^2 + A F^2 - 2 A E cdot A F cos alpha ] - Substituting the values, we get: [ E F^2 = (8 R sin beta sin gamma)^2 left[ sin^2 left( frac{pi}{3} + beta right) + sin^2 left( frac{pi}{3} + gamma right) - 2 sinleft(frac{pi}{3} + betaright) sinleft(frac{pi}{3} + gamma right) cos alpha right] ] 13. **Using Trigonometric Identity**: - We use the identity: [ sin^2 alpha = sin^2 left( frac{pi}{3} + beta right) + sin^2 left( frac{pi}{3} + gamma right) - 2 sin left( frac{pi}{3} + beta right) sin left( frac{pi}{3} + gamma right) cos alpha ] - Thus, [ E F = 8 R sin alpha sin beta sin gamma ] # Conclusion: [ boxed{E F = 8 R sin alpha sin beta sin gamma} ]

answer:1. Start with the given equation: [ frac{x}{x-1} = frac{y^2 + 3y + 2}{y^2 + 3y - 1} ] 2. Cross multiply to clear the denominators: [ x(y^2 + 3y - 1) = (x-1)(y^2 + 3y + 2) ] 3. Expand and arrange both sides: [ xy^2 + 3xy - x = xy^2 + 3xy + 2x - y^2 - 3y - 2 ] 4. Simplifying the equation by canceling and rearranging terms: [ -x - 2x = -y^2 - 3y - 2 ] [ -3x = -y^2 - 3y - 2 ] [ x = frac{y^2 + 3y + 2}{3} ] Conclusion: (frac{y^2 + 3y + 2{3}}) is the value of (x) that satisfies the equation. The final answer is boxed{A) (frac{y^2 + 3y + 2}{3})}