answer:We are given the equation: [x^{2} + 2x + 2|x+1| = a] and we need to find the values of the parameter (a), for which this equation has exactly two roots. To solve this, we start by considering the two cases for (|x + 1|): 1. **Case 1: (x geq -1)** In this case, (|x+1| = x+1). Substituting into the equation yields: [ x^{2} + 2x + 2(x+1) = a ] Simplifying, we get: [ x^{2} + 2x + 2x + 2 = a ] [ x^{2} + 4x + 2 = a ] [ x^{2} + 4x + 2 - a = 0 ] This is a quadratic equation in the form (Ax^2 + Bx + C = 0), where (A = 1), (B = 4), and (C = 2-a). 2. **Case 2: (x < -1)** In this case, (|x+1| = -(x+1)). Substituting into the equation yields: [ x^{2} + 2x + 2(-(x+1)) = a ] Simplifying, we get: [ x^{2} + 2x - 2x - 2 = a ] [ x^{2} - 2 = a ] [ x^{2} - a = 2 ] [ x^{2} = a + 2 ] Here, we solve for (x): [ x = pm sqrt{a + 2} ] For the quadratic equation in **Case 1** to have exactly one root, its discriminant must be zero. - The discriminant (Delta) of (x^2 + 4x + (2 - a) = 0) is given by: [ Delta = B^2 - 4AC ] [ Delta = 4^2 - 4 cdot 1 cdot (2 - a) ] [ Delta = 16 - 4(2 - a) ] [ Delta = 16 - 8 + 4a ] [ Delta = 8 + 4a ] For exactly two roots overall, this discriminant must be greater than zero and not equal, ensuring distinct roots on both ends. So we have: [ 8 + 4a = 0 ] [ 4a = -8 ] [ a = -2 ] Moving forward, ensuring there is exactly one root, for the mix of positive and negative: Upon further analyzing: Conclusion: For there to be roots separated and inclusive of: [ a > -1 ] Thus, ensuring there are two distinct roots providing: Thus, the required value of the parameter (a) is: [ boxed{a > -1} ]

answer:Given the problem scenario, we need to analyze the times T_{1}, T_{2}, and T_{3} for the three sprinters A, B, and C respectively, based on their speeds and how they ran the 100m race. Let's proceed step by step: 1. **Sprinter A's Time (T_{1}):** Sprinter A ran half of the time at a speed of V_{1} meters per second and the other half at V_{2} meters per second. The average speed for A can be calculated using the harmonic mean of V_{1} and V_{2} because the time is evenly split between the two speeds. Thus, the average speed of A is frac{2V_{1}V_{2}}{V_{1} + V_{2}} meters per second. Therefore, the time taken by A, T_{1}, can be calculated as: [ T_{1} = frac{100}{frac{V_{1} + V_{2}}{2}} ] 2. **Sprinter B's Time (T_{2}):** Sprinter B ran the entire race at a constant speed of sqrt{V_{1}V_{2}} meters per second. Thus, the time taken by B, T_{2}, is: [ T_{2} = frac{100}{sqrt{V_{1}V_{2}}} ] 3. **Sprinter C's Time (T_{3}):** Sprinter C ran half of the distance at V_{1} meters per second and the other half at V_{2} meters per second. The effective speed for C is the harmonic mean of V_{1} and V_{2} due to the distance being split. Therefore, the time taken by C, T_{3}, is: [ T_{3} = frac{100}{frac{2V_{1}V_{2}}{V_{1} + V_{2}}} ] 4. **Analyzing the Results:** - We know from the inequality of arithmetic and geometric means that frac{V_{1}+V_{2}}{2} geqslant sqrt{V_{1}V_{2}} geqslant frac{2V_{1}V_{2}}{V_{1}+V_{2}}. Since time is inversely proportional to speed, we can deduce that T_{1} leqslant T_{2} leqslant T_{3}. - The equality holds only when V_{1} = V_{2}, confirming choice A as correct and B as incorrect. - Additionally, since frac{V_{1} + V_{2}}{2} cdot frac{2V_{1}V_{2}}{V_{1} + V_{2}} = (sqrt{V_{1}V_{2}})^2, we can confirm that {T}_{1}{T}_{3}={T}_{2}{}^{2}, making choice C correct. - Lastly, the relationship frac{1}{{T}_{1}}+frac{1}{{T}_{3}} = frac{2}{{T}_{2}} does not hold based on our calculations, making choice D incorrect. Therefore, the correct conclusions are boxed{A} and boxed{C}.

answer:Let's solve the expression step by step: First, we'll solve the innermost parentheses and exponents: (28 * 2 + (48 / 6) ^ 2 - 5) * (69 / 3) + 24 * (3 ^ 2 - 2) = (28 * 2 + (8) ^ 2 - 5) * (69 / 3) + 24 * (9 - 2) Now, we'll calculate the squares and the multiplications/divisions: = (56 + 64 - 5) * (23) + 24 * (7) = (120 - 5) * 23 + 168 = 115 * 23 + 168 Now, we'll do the multiplication: = 2645 + 168 Finally, we'll add the last two numbers: = 2813 So the value of the expression in approximate form is boxed{2813} .

answer:1. **Define the fractions and their properties:** Let the two fractions be frac{a}{600} and frac{b}{700} where gcd(a,600)=1 and gcd(b,700)=1. This means that a and b are coprime with their respective denominators. 2. **Sum the fractions:** [ frac{a}{600} + frac{b}{700} = frac{700a + 600b}{600 times 700} = frac{700a + 600b}{420000} ] Simplify the numerator and denominator: [ frac{700a + 600b}{420000} = frac{7a + 6b}{4200} ] 3. **Factorize the denominator:** [ 4200 = 2^3 times 3 times 5^2 times 7 ] 4. **Determine the conditions for the numerator:** We need to find the minimum value of the denominator of the irreducible fraction frac{7a + 6b}{4200}. For this, we need to ensure that gcd(7a + 6b, 4200) is as large as possible. 5. **Use Bezout's Lemma:** By Bezout's Lemma, any integer can be expressed as 7a + 6b for some integers a and b. Let 7a + 6b = n, where n is an integer. We need to find the conditions under which n is coprime with the factors of 4200. 6. **Analyze the gcd conditions:** Since gcd(a, 600) = 1 and gcd(b, 700) = 1, it follows that gcd(n, 6) = 1 and gcd(n, 7) = 1. This means that n cannot be divisible by 2, 3, or 7. 7. **Determine the maximum gcd:** The maximum possible gcd of 4200 and n is 25, since n must be coprime with 2, 3, and 7. Therefore, the minimum denominator of frac{7a + 6b}{4200} = frac{n}{4200} is: [ frac{4200}{25} = 168 ] 8. **Verify the solution:** A denominator of 168 is satisfied by choosing a = 25 and b = -25, since: [ 7a + 6b = 7(25) + 6(-25) = 175 - 150 = 25 ] Thus, gcd(25, 4200) = 25, and the fraction reduces to frac{25}{4200} = frac{1}{168}. The final answer is boxed{168}.