[high school physics] the maximum speed of a car with rated power of 80kW on a straight road is 20m / s, and its mass m = 2 × 10 ^ 3kg If the car starts to make uniform acceleration linear motion from static, the acceleration is 2m / S ^ 2 (1) Constant resistance of a car (2) Instantaneous power of automobile at the end of 3S (3) How long does it take for the car to reach its rated power?

[high school physics] the maximum speed of a car with rated power of 80kW on a straight road is 20m / s, and its mass m = 2 × 10 ^ 3kg If the car starts to make uniform acceleration linear motion from static, the acceleration is 2m / S ^ 2 (1) Constant resistance of a car (2) Instantaneous power of automobile at the end of 3S (3) How long does it take for the car to reach its rated power?

The maximum speed of a car with rated power of 80kW on a straight road is 20m / s, and the mass of the car is 2T. If the car starts to move in a straight line with uniform acceleration from a standstill, the acceleration is 2m / S ^ 2, and the resistance in the process of motion remains unchanged, we can find: (1) what is the resistance of the car? (2) what is the instantaneous power of the car at the end of 3S

It is known that (x2-x + 1) 6 = a12x12 + a11x11 + +A2x2 + a1x + A0, then A12 + A10 + A8 + +a2+a0=______ ．

∵（x2-x+1）6=a12x12+a11x11+… +A2x2 + a1x + A0, when x = 1: ∵ (x2-x + 1) 6 = A12 + a11 + +When x = - 1, (x2-x + 1) 6 = a12-a11 + +a2-a1+a0=729，②∴①+②=2（a12+a10+a8+a6+a4+a2+a0）=730，∴a12+a10+a8+...

（x²-x+1)6=a12x12+a11x11+… +A1x + A0, then A12 + a11 + a1=
I checked the answers of several websites, and we all use x = 1, - 1 to substitute and ask again. Why can we do that?

Because left and right are connected by an equal sign, when you take x = 1 and - 1, it still satisfies the equation

If the sixth power of (x to the second power + x-1) = a12x to the 12th power +. A2x to the second power + A to the first power X + A0, then A12 + A10 +. A8... + A0
Then A12 + A10 +. A8... + A0 is equal to?

Let x = - 1
Then the 12th power of a12x + the 11th power of a11x + +The second power of a2x + a1x + A0
=A12-A11+A10-…… +A2-A1+A0
So a12-a11 + a10 - +A2-A1+A0
=[(-1)²-(-1)+1]^6
=729
And A12 + a11 + A10 + +A2 + A1 + A0 = (1 & # 178; - 1 + 1) ^ 6 = 1
2(A12+A10+A8+A6+A4+A2+A0)=730
A12+A10+A8+A6+A4+A2+A0=365