If the quadratic polynomial x2 + 2kx-3k2 can be divided by X-1, then the value of K is______ ．

If the quadratic polynomial x2 + 2kx-3k2 can be divided by X-1, then the value of K is______ ．

∵ the polynomial x2 + 2kx-3k2 can be divided by X-1, ∵ X-1 is the factor of the polynomial x2 + 2kx-3k2, ∵ X-1 = 0, that is, x = 1 is the solution of the polynomial x2 + 2kx-3k2, ∵ 12 + 2k-3k2 = 0, solving the univariate quadratic equation about K, we get K1 = 1, K2 = - 13, so the answer is: k = 1 or K = - 13

If the quadratic polynomial x2 + 2kx-3k2 can be divided by X-1, then the value of K is______ ．

∵ the polynomial x2 + 2kx-3k2 can be divided by X-1, ∵ X-1 is the factor of the polynomial x2 + 2kx-3k2, ∵ X-1 = 0, that is, x = 1 is the solution of the polynomial x2 + 2kx-3k2, ∵ 12 + 2k-3k2 = 0, solving the univariate quadratic equation about K, we get K1 = 1, K2 = - 13, so the answer is: k = 1 or K = - 13

If the square + 2x + k of quadratic polynomial x can be divided by X-1, try to find the value of K

The square + 2x + k of quadratic polynomial x can be divisible by X-1
It is shown that the square + 2x + k of quadratic polynomial X has the factor X-1
Furthermore, when x = 1, the square of quadratic polynomial x + 2x + k = 0
So 1 + 2 + k = 0
k=-3

A and B start from AB and go opposite each other. For the first time, they meet 80 kilometers away from a, and then continue to move on,
After arriving at the starting point of the other party, return immediately, and meet for the second time at a distance of 70 km from B. find the distance between AB and B?

In the first meeting, the two sides took a total of one journey. In the second meeting, the two sides took a total of three journeys since they started. Because they all travel at a constant speed, for car a (or car B), the distance from starting to the second meeting is three times that of the first meeting
S+70=3*80,S=170

Rational numbers are classified according to their properties
① Classification by definition
② Classification by the nature of numbers

① Classification by definition
{integer: {positive integer, 0, negative integer
Score: {positive score, negative score
② Classification by the nature of numbers
{positive rational number: positive integer, positive fraction
0
Negative rational number: negative integer, negative fraction

A. B is 1000km away from each other, and a and B trains
A. The distance between B and a is 1000km. A and B trains start at the same time from A. B. they meet on the way. A car arrives at B 15 hours after meeting, and B car arrives at a 6 and 2 / 3 hours after meeting. If the speed of a car is 1.5 times that of a car, the speed of a and B cars should be calculated

Set the speed of a and B as V1 and V2 unit km / h respectively
From the theme
After meeting, the distance a runs is 15 * v1
The running distance of B is 20 / 3 * v2
The sum of the two is ab, that is 1000;
Two equations
15*v1+20*V2/3=1000
V2=1.5*V1
Joint solution
V1 = 40 km / h
V2 = 60 km / h

Average the data of the following groups: (1) 1,3,7,9,12,13,17,24; (2) 28,29,30,31,32,33,34; (3) 4,5,7,7,7,6,6,4,4,7

1, the average is 86 / 8 = 10.75
2, each number minus 31, get a new group of numbers: - 3, - 2, - 1,0,1,2,3. Because the average of the new group of numbers is 0, the average of the original data is 31
3, average = (4 × 3 + 7 × 4 + 6 × 2 + 5) / 10 = 57 / 10 = 5.7

A car runs 140 kilometers in 2 hours. At this speed, it runs 5 hours from a to B. how long is the highway between a and B?

A: the highway between a and B is 350 km long

In the 21st century, the number of years in some years is composed of four different numbers. There are? Years in total

20: The last two numbers can be any two of the eight numbers 1, 3, 5, 4, 5, 6, 7, 8 and 9. The first number can be any one of the eight numbers, and the second number can be any one of the remaining seven numbers, with 8 * 7 = 56

A and B leave from a and B at the same time and meet in 2 hours. After meeting, the two vehicles continue to move forward. When a arrives at B, the two vehicles stop at each other

A and B leave from a and B at the same time. They meet in 2 hours. After meeting, they continue to move forward. When a arrives at B, B is 60 kilometers away from A. It is known that the speed ratio of the two vehicles is 3:2
The solution is as follows:
1. Suppose the speed of car a and car B is 3x and 2x respectively;
2. Suppose the meeting point is C, then the distance is AC = 6x, BC = 4x; (according to the known condition "a and B drive relatively and meet in 2 hours", that is, the distance AC and BC that a and B can walk for 2 hours respectively);
3. After the meeting, the time it takes for car a to reach the destination after walking BC section is 4-3 hours (distance bc-car speed;
4. The distance that car B can walk for 4-3 hours is (4-3) * 2x; it has to walk 60 kilometers more to reach point a;
5. Using the fourth point series equation: (4 / 3) * 2x + 60 = 6x, the solution is x = 18;
6. From the first point, we can get the speeds of a and B vehicles: 3 * 18 = 54; 2 * 18 = 36; 3 * 18 = 54; 2 * 18 = 36;
Let the speed of a and B be 3x2x respectively;
（4\3)*2x+60＝6x
The solution is: x = 18
The speed of car a is: 3x = 3 * 18 = 54
Speed of car B: 2x = 2 * 18 = 36
A: the speed of car a is 54 km / h, and that of car B is 36 km / h;