A + B + C = 0 ABC = 2 find the value of a B C
Is the condition C = - a-b - AB (a + b) = 2 insufficient?
RELATED INFORMATIONS
- 1. Given that ABC is a positive integer and satisfies a + C = 10, C + B = 13, find a, B,
- 2. Given that ABC is a positive integer and satisfies a + C = 10, C + B = 13, try to judge whether a, B and C can form a triangle
- 3. Given that a, B and C are all positive integers, and ABC = 2008, then the minimum value of a + B + C is______ .
- 4. Given that a, B and C are all positive integers, and ABC = 2008, then the minimum value of a + B + C is______ .
- 5. Given that a, B and C are all positive integers, and ABC = 2008, then the minimum value of a + B + C is______ .
- 6. a. B. C is three positive integers, known ABC = 2008, find the minimum value of a + B + C What is this mathematical idea
- 7. In △ ABC, the lengths of three sides are positive integers a, B and C respectively, and C ≥ B ≥ a > 0. If B = 4, then such triangles have the same length______ One
- 8. It is known that the lengths of a, B and C on three sides of a triangle ABC are integers, and a is less than or equal to B and less than or equal to C. If b = m (m ∈ positive integer), then such a triangle has I know the answer is m (M + 2) / 2, but I don't know why There are several (in M)
- 9. In the triangle ABC, if three sides are known to be continuous positive integers, The cosine value of the maximum angle is - 1 / 4. Find the maximum area of the parallelogram with the maximum angle as the inner angle and the sum of the two sides of the angle as 4
- 10. If a, B and C are all positive numbers, then (a + b) (B + C) (c + a) > = 8abc
- 11. A + B + C = 0, ABC finds a (B + C) + B (a + C) + C (a + b) ≠ abc≠0
- 12. Given that a + B + C = 0 and ABC ≠ 0, the value of a (1 △ B + 1 △ C) + B (1 △ C + 1 △ a) + C (1 △ a + 1 △ b) + 3 is calculated
- 13. It is known that a, B and C satisfy a + B + C = 0 and ABC = 8. It is proved that 1 / A + 1 / B + 1 / C is less than 0
- 14. Given a > 0, b > 0, and ABC = 1, prove (1 + a) (1 + b) (1 + C) 8
- 15. Given a + B + C = 0, ABC = 8, it is proved that 1 / A + 1 / B + 1 / C is less than 0
- 16. Given a + B + C = 0, ABC = 8, 1 / A + 1 / B + 1 / C
- 17. a. B, C ∈ R +, and (1 + a) (1 + b) (1 + C) = 8
- 18. If a, B, C > 0. ABC = 8
- 19. Given that a, B, C > 0 and 1 / (1 + a) + 1 / (1 + b) + 1 / (1 + C) = 1, we prove that ABC ≥ 8
- 20. It is known that the lengths of three sides of the triangle ABC are a, B, C respectively, and a & gt; C, then | C-A | - √ (AC-B) & # 178=