Given a = 2010, B = 2012, find quarter a & # 178; - half AB + quarter B & # 178;
Quarter a and 178; - half AB + quarter B and 178;
=Quarter (A & # 178; - 2Ab + B & # 178;)
=One quarter (a-b) & 178;
=Quarter (2010-2012) & 178;
=Quarter (- 2) & 178;
=Quarter * 4
=1
RELATED INFORMATIONS
- 1. In the indefinite integral of higher mathematics, is the order of u determined in the method of integration by parts against the power exponent three or against the power exponent three Please give an example to explain whether "three" can reach "Zhi", or "Zhi" can reach "three",
- 2. It is fast to find the indefinite integral Thank you Wrong. It's the power of the square of e ^ (- t)
- 3. If A-B = 2, B-C = 1, then a & # 178; + B & # 178; + C & # 178; - AB BC AC= Hurry!
- 4. Calculation: (B-C) / (A & # 178; - AB AC + BC) - (C-A) / (B & # 178; - BC AB + AC) + (a-b) / (C & # 178; - AB BC + AB)
- 5. Calculation: (B-C) / (A & # 178; - AB AC + BC) - (C-A) / (B & # 178; - BC AB + AC) + (a-b) / (C & # 178; - AC BC + AB)
- 6. Given a + B + C = 4, a & # 178; + B & # 178; + C & # 178; = 6, find AB + BC + AC =?
- 7. Prove: A & # 178; + B & # 178; + C & # 178; ≥ AB + BC + AC
- 8. If a and B satisfy the absolute value of a + B-7 + (AB + Half) 178; = 0, find the value of 2 (a + b) + ab - (a + b) - 3AB
- 9. If x2 + X-1 = 0, then the value of the algebraic expression X3 + 2x2-7 is () A. 6B. 8C. -6D. -8
- 10. If AB is opposite to each other and CD is reciprocal to each other, then the algebraic formula (a + b) & # 178; - (CD) is 1 / 100
- 11. Given that the quadratic function f (x) satisfies the conditions f (0) = 1, f (x + 1) - f (x) = 2x, the number of solutions of F (| x |) = a, a belonging to R is discussed
- 12. It is known that the quadratic function f (x) satisfies f (x + 1) - f (x) = 2x (x ∈ R), and f (0) = 1 It is known that f (x) is a quadratic function. For any x belonging to R, f (x + 1) - f (x) = - 2x + 1 and f (0) = 1 are satisfied (1) Finding the analytic expression of F (x) (2) When x belongs to [- 2,1], the image of y = f (x) is always above the image of F = - x + m, and the value range of real number m is obtained I would like to ask the following constant set up into a solution Is m ∈ [- 1,5] the answer? (2) In the question, f (x) = 2x + m has a solution, and the range of real number m is obtained
- 13. It is known that f (1-x) = f (1 + x) holds for any quadratic function f (x) belonging to R Let a = (SiNx, 2), B = (2sinx, 1 / 2), C = (cosx, 1), d = (1,2). When x belongs to [0, π], find the solution set of the inequality f (a × b) > F (C × d)
- 14. It is known that the coefficient of quadratic term of quadratic function f (x) is negative, and for any x, it is constant and f (2-x) = f (2 + x) holds. Solve the inequality f [Log1 / 2 (x ^ 2 + X + 1 / 2)]
- 15. If f (x) is a quadratic function and f (0) = 1, f (2x + 1) - f (x) = 2, find f (x)
- 16. (1) If f (x) is a quadratic function, and f (0) = 1, f (x + 1) - f (x) = 2x, find (x) (2) The value range of y = radical (x2-16) is
- 17. Let the minimum value of quadratic function y = f (x) be 4, and f (0) = f (2) = 6
- 18. It is known that the quadratic function f [x] satisfies f [2-x] = f [2 + x], and the intercept of the image on the y-axis is 0, and the minimum value is negative one
- 19. Given the quadratic function y = f (x), satisfying f (- 2) = f (0) = 0, and the minimum value of F (x) is - 1. (1) if the function y = f (x), X ∈ R is odd, when x > 0, f (x) = f (x), find the analytic expression of the function y = f (x), X ∈ R; (2) Let G (x) = f (- x) - λ f (x) + 1, if G (x) is a decreasing function on [- 1, 1], find the value range of real number λ
- 20. It is known that the quadratic function y = f (x) satisfies f (- 2) = f (0) = 0, and the minimum value of F (x) is - 1 (1) If the function y = f (x), X ∈ R is an odd function, when x > 0, f (x) = f (x), find the analytic expression of the function y = f (x), X ∈ R (2) Let g (x) = f (- x) - t · f (x) + 1. If G (x) is a decreasing function on [- 1,1], the value range of real number T is obtained