Given that the square of x minus 3 times the square of Y equals 2XY, x > 0, Y > 0, then the value of (x + 2Y) / (X-Y) is?

Given that the square of x minus 3 times the square of Y equals 2XY, x > 0, Y > 0, then the value of (x + 2Y) / (X-Y) is?

When x > y, the answer is 5 / 2, when XY, X-Y = 2Y, when x

If x0, find the value of LXL / x + Lyl / y + lxyl / XY

lxl/x+lyl/y+lxyl/xy
=-1+1-1
=-1

A car goes from city a to city B in four fifths of an hour, three tenths of the whole journey. According to this calculation, what is the percentage of the journey in two hours?

Three tenths divided by four fifths multiplied by two

What number does the letter a that meets the condition | a | / a = - 1 represent?

A is a negative number

The maximum acceleration of a fighter on an aircraft carrier during takeoff is 4.5m/s,
The maximum acceleration of a fighter on an aircraft carrier during take-off is a = 4.5m/s, and the aircraft can only take off when the speed reaches V0 = 60m / s. the deck length of the aircraft carrier is L = 289m. In order to make the aircraft take-off safely, the aircraft carrier should sail at a top speed to ensure the safety of take-off. What is the minimum speed V of the aircraft carrier? (assuming that the aircraft take-off has no effect on the state of the aircraft carrier, The motion of an aircraft can be regarded as uniformly accelerated linear motion
Taking the aircraft carrier as the reference frame, the relative initial velocity of the fighter is 0, the relative final velocity is V1 = V0-V, the relative acceleration is still a, and the relative displacement is L,
Research aircraft: from the beginning to take-off process, known by the laws of kinematics
2aL=v1^2
Substituting the data, the solution is V0 = 9 m / s
Since it is based on the aircraft carrier as the reference frame, the initial speed of the fighter is 0. Then the final speed should be v0. How can it be V0-V,

V0 is relative to the ground

The difference of two x plus five times twenty minus x equals sixty-four solutions

2x+5.(20-x)==64
-3x=-36
x=12

Traction problems of automobile passing through arch bridge
When a car runs on an arch bridge at a constant speed, it receives the centripetal force provided by the resultant force of gravity and supporting force at the highest point. Why do we ignore the traction force and friction force? When we do not analyze the force at the highest point, we should consider the traction force. Why?
Clear analysis!

The centripetal force points to the center of the circle, while the traction force and the friction force are in the tangent direction. These two forces have no contribution to the centripetal force, which means they have no contribution to the centripetal acceleration

5x + 4 + 2 equals 8, how much is x

5x + 4 = 6 -- 5x + 4 = 6 or 5x + 4 = - 6.. so x = - 2 or x = 0.4

The distance between a and B is 40km. A starts for 1.5h and then B starts. A is behind and B is in front. They walk in the same direction. The speed of a is 8km / h and that of B is 6km / h. how many hours after a start, can a catch up with B?

Let a catch up with B after starting for X hours. From the meaning of the question: 8x-6 (x-1.5) = 40, the solution is x = 15.5. Answer: a catch up with B after starting for 15.5 hours

If a + B = 1, the second power of a + the second power of B = 2, then the value of AB is?
There are four options: a - 1 B3 C - 3 / 2 D - 1 / 2

Because a + B = 1, then (a + b) ^ 2 = 1, so a ^ 2 + 2Ab + B ^ 2 = 1 (1)
And a ^ 2 + B ^ 2 = 2 (2)
So (1) - (2) get 2Ab = - 1
ab=-1/2
So choose D