2A ^ - 6 factoring in real numbers

RELATED INFORMATIONS

• 1. (2a + b) (2a-3b) - 3A (2a + b) factorization
• 2. 3A ^ 2-10 factoring in real numbers
• 3. 1. Decompose factor A ^ 4-3a ^ 2 + 2 in the range of real number 2. Try to prove: no matter m takes any real value, the value of the algebraic formula 3M ^ 2 + 4m + 5 is always greater than zero
• 4. -AB ^ 2 + 2A ^ 2b-a ^ 3B factorization
• 5. Factorization: - AB + 2A ^ 2b-a ^ 3B
• 6. Factorization factor AB ^ 2-2a ^ 2B + A ^ 3 The square of ab-the square of 2A × the third power of B + 3
• 7. Decomposition factor: 2a2b-a3-ab2
• 8. Decomposition factor A ^ 3 + AB ^ 2-2a ^ 2B,
• 9. Given a + B = 5, ab = 3, find the value of one part of a + one part of B
• 10. If we know / A / = 5, B = 3 and ab < 0, we can find the value of a + B
• 11. 3a-5b-2a + B simplification
• 12. 1. First simplify, then evaluate a - {b-2a + [3a-2 (B + 2a) + 5B]}, where a = 2009, B = 2010 2. Given X & sup2; + xy = 2, Y & sup2; + xy = 5, what is the value of 1 / 2x & sup2; + XY + 1 / 2Y & sup2? 3. If - 0.2A (3x power) B & sup3; and 1 / 2A & sup2; B (y power) are of the same kind, Find the value of the algebraic formula 3xy & sup2; - [2x & sup2; Y-2 (xy-3 / 2XY & sup2;) + XY] + 3x & sup2; y
• 13. First simplify, then evaluate: (A-1) / a △ {a - (2a-1) / a}, where a = √ 3 + 1 First simplify, then evaluate: (A-1) / a △ {a - (2a-1) / a}, where a = √ 3 + 1
• 14. Simplify evaluation 3 (a + 1) ^ 2 - (2a + 1) (2a-1), where a = 1
• 15. Simplified evaluation: 3 (a + 1) ^ 2 - (a + 1) (2a-1) where a = 1 It's simplified evaluation!
• 16. First simplify and then evaluate: (2a + 1) ^ 2 - (2a + 1) (2a-1), where a = change sign 5-1 / 2
• 17. First simplify, then evaluate: 1 / (a + 1) - (a + 3) / (a ^ 2-1) * (a ^ 2-2a + 1) / (a ^ 2 + 4A + 3), where a satisfies a ^ 2 + 2a-1 = 0
• 18. First simplify, then evaluate A-1 / A + 2 × a ^ 2-4 / A ^ 2-2a + 1 △ 1 / A ^ 2-1, where a satisfies a ^ 2-A = 0 (A-1 / A + 2) × (a ^ 2-4 / A ^ 2-2a + 1) / (1 / A ^ 2-1), where a satisfies a ^ 2-A = 0
• 19. Simplification: (3a-2b) * (2a + 5b) =?
• 20. First simplify, then evaluate 3A & sup2; B - [2Ab & sup2; - 2 (ab-3 / 2 A & sup2; b)] + 2Ab 3A & sup2; B - [2Ab & sup2; - 2 (ab-3 / 2 A & sup2; b)] + 2Ab. Where a and B satisfy │ B + 1 │ + (2a-4) & sup2; = 0 (detailed process is required)