If A-B = 5, the square of a + the square of B = 17, then AB = a + B=

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• 1. (1 + 2 + 3 + 4 +... + 98 + 99 + 100) / 5 is equal to?
• 2. Mathematical problem of senior one: it is known that y = f (x) is an increasing function in the domain (- 1,1) and f (1 + a) is less than f (A2-1) (A2 represents the square of a, then the range of a is
• 3. 1. Given that the domain of F (x + 1) is [- 1,1], then the domain of F | x + 1 | is the process and method of? - (3,1) 2. Given the function f (x) = 1 / 2x ^ 2-x + 3 / 2, whether there is a real number m, so that the domain of definition and range of value of the function are [1, M] (M > 1)? If there is, request m m = - 3, 1 / 2x ^ 2 is the square of half X
• 4. The domain of F (x + 3) is [2,3], (1) the domain of F (x); (2) the domain of F (X-2)
• 5. How much is the 3 / 2 power of 8 / 27? Is there a - 3 / 2 power of 8 / 27? How much is it? For example, please give the explanation and answer of the problem. What is the value of (1-0.5 to the power of - 2)?
• 6. The power a of (3 / 2) multiplied by the power B of (8 / 27) De is equal to 2 / 3, and the value can be obtained from a-2b + 2
• 7. Given a + B = 7, ab = 5, find the value of (a-b) square
• 8. Given a + B = 5, ab = 7, find the square of a, B + the square of ab-a-b
• 9. Formula: (one half of the square of AB) third power
• 10. Given the quadratic power of a + AB = 10, the quadratic power of B - AB = 7, find the quadratic power of a + 2ab-b
• 11. If the three sides a, B and C of a triangle are suitable for A2 (B-C) + B2 (C-A) + C2 (a-b) = 0, then the shape of △ ABC is () A. Right triangle B. isosceles triangle C. isosceles right triangle D. equilateral triangle
• 12. If three sides a, B and C of triangle ABC satisfy (a-b): (C-B): (a + b) = 7:1:18, the shape of the triangle is judged There is a reasoning process
• 13. The length of the three sides of a triangle is ABC. If a and B are equal to 10, AB is equal to 18, and C is equal to 8, judge the shape of the triangle,
• 14. It is known that a, B and C are the three sides of △ ABC. It is proved that | a + B-C | + | a-b-c | = 2B Such as the title
• 15. In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10, C = √ In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 C = √ 5 for ABC area
• 16. In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 What's the math problem In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 Find the value of a + B?
• 17. In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, and satisfy b square + C square - a square = BC 1. Find the value of angle a 2. Find the value of sin (a + 15 degrees)
• 18. In the triangle ABC, A.B.C is the three sides opposite to the angle A.B.C, a square - (B-C) square = BC, 1. Find the angle A 2. If BC = 2 times root sign 3, angle B is equal to X and perimeter is y, find the value range of function y = f (x)
• 19. Given that A.B.C is a positive integer, what is the answer to prove that (a + b) (B + C) (c + a) is greater than or equal to 8abc
• 20. (a + b) * (B + C) * (a + C) greater than or equal to 8abc