The following inequalities are proved: (1) if a > b and C < 0, then (a-b) C < 0 (2) If a < B < 0, then 0 > 1 / a > 1 / b

The following inequalities are proved: (1) if a > b and C < 0, then (a-b) C < 0 (2) If a < B < 0, then 0 > 1 / a > 1 / b

(1) If a > b and C < 0, then (a-b) C < 0
From the known, we get: A-B > 0, C0, divide each side of the known inequality by a positive number, and the inequality direction remains unchanged
a/(ab)1/b

Prove the inequality: A.B.C ∈ R, a ^ 4 + B ^ 4 + C ^ 4 ≥ ABC (a + B + C)

a^4+b^4>=2a^2*b^2
a^4+c^4>=2a^2*c^2
2a^4+b^4+c^4>=4a^2*bc
Similarly, 2b ^ 4 + C ^ 4 + A ^ 4 > = 4AB ^ 2 * C
2c^4+a^4+b^4>=4abc^2
Add
4a^4+4b^4+4c^4>=4a^2*bc+4ab^2*c+4abc^2
That is, a ^ 4 + B ^ 4 + C ^ 4 > = ABC (a + B + C)
Obtain the equal sign when a = b = C

Known function f (x) = cos (2x + π / 3) + sin ² x; X find the minimum positive period and maximum value of F (x)

F (x) = 1 / 2cos2x radical 3 / 2sin2x + 1 / 2 (2Sin ^ 2x-1 + 1)
=-Root 3 / 2sin2x + 1 / 2
The maximum value is (1 + radical 3) / 2
The minimum positive period is 2 π / 2 = π

When the displacement is 500m, it takes 15s. I don't know how to find the motion relationship by calculus

Acceleration is the change of velocity per unit time. This is the definition. The velocity at any time is (vcho + at), and the differential of time is DT. There are: DS = (vcho + at) * DT (the time is very short, and the change of velocity is very small, which can be calculated by the formula of uniform motion). Integrate both sides at the same time: S = ∫ (vcho + at) * DT = vcho T + 1 / 2 * a

The particle with mass m makes a uniform circular motion around the circular orbit of radius r at speed V, and the resultant impulse is in half a cycle A. 0 b.mv c.2mv D. conditions are insufficient to determine

It can be solved by the momentum theorem. The velocity remains the same and the direction is reversed, so the resultant impulse is 2mV. Choose C
See picture:

A calculus physics problem, urgent 1. There is a wooden stick with length L and mass M. Rao rotates vertically through a rotating axis with angular velocity W. calculate the centripetal force by integral method

It is impossible to use the integral method. According to Newton's law of motion and the laws of motion of the center of mass, the centripetal acceleration can be obtained by the * differential * method as l * W ^ 2 / 2 (if the rod is uniform and the center of mass is in the center). Therefore, the centripetal force on the rod is L * W ^ 2 / 2 * M
Using this formula, minus the component of gravity in the direction of the rod, you get the centripetal force of the shaft to the rod