If the quadratic polynomial x + 2kx-3k can be divided by X-1, try to find the value of K

If the quadratic polynomial x + 2kx-3k can be divided by X-1, try to find the value of K

If x + 2kx-3k = (x-1k) (x + 3K) can be divided by X-1, then x-1k = X-1, or x + 3K = X-1

Given that x2 + X-6 is a factor of the polynomial 2x4 + x3-ax2 + BX + A + B-1, then a=______ ；b=______ ．

Let another factor be: 2x2 + MX + N, then (x2 + X-6) (2x2 + MX + n) = 2x4 + (M + 2) X3 + (M + n-12) x2 + (n-6m) x-6n, then: M + 2 = 1m + n − 12 = − an − 6m = Ba + B − 1 = − 6N, the solution is: M = − 1n = − 3A = 16b = 3, so the answer is: 16, 3

In the quadrilateral ABCD, it is known that the diagonal AC intersects BD, and AB, BC, ad, DC (or extension) intersect plane α at e, F, G, h respectively,
Prove that e, F, G and H are collinear

∵ quadrilateral ABCD is a plane quadrilateral
The AB, BC, ad, DC (or extension line) are coplanar
Let this plane be β, and the intersection of α and β be L
∵ e, F, G, h are both in α and β
E, F, G and H are on the intersection l of α and β, that is, four points are collinear

For a pile of coal, 20% will be used in the first time, and 6 tons will be used in the second time, leaving 14 tons. How many pairs of coal are there
For a pile of coal, 20% will be used in the first time, and 6 tons will be used in the second time. There are still 14 tons left. How many tons of coal are there?
Equation or whatever, help me solve it as soon as possible!

20% is used for the first time, and the remaining 80% corresponds to a weight of 6 + 14 = 20 tons,
Originally 20 △ 80% = 25 (tons)

When can't mean inequality be used to solve the maximum value of check function?
Many people say that we should use the mean inequality to solve the function. But I remember that all the problems we have done are that the mean inequality is not applicable, but the type of the problem has been forgotten by me. So let's ask when we can use the mean inequality and when we can't?

There is no possibility that is not applicable, but sometimes you can't take the equal sign, such as (x ^ 2 + 2) + 1 / (x ^ 2 + 2) let x ^ 2 + 2 = t, change to t + 1 / T ≥ 2, the equal sign is obtained when x ^ 2 + 2 = 1 / (x ^ 2 + 2), that is, (x ^ 2 + 2) ^ 2 = 1, that is, you need x ^ 2 + 2 = 1, but because x ^ 2 > 0, so x ^ 2 + 2 can't get 1, only get 2, and the check function is greater than

A batch of goods will be distributed to Party A and Party B according to the ratio of 5:3. Party A will transport four fifths of the task of the team, and the rest will be transported to Party B. Party B will transport a total of 480 goods

Team a should complete 480 / (4 / 5) = 600, then team B actually completes 600 * 3 / 5 + 600-480 = 480
Hope to adopt it!

It is known that the circle C is tangent to both axes, and the distance from the center of the circle C to the straight line y = - x is equal to
2．
(1) Find the equation of circle C
(2) To find the minimum value of AOB area of triangle, I think only the maximum value can be calculated. When s = 1 / 2Mn, M = n, take the maximum value. What's wrong with me
The problem is that NX + my Mn = 0 is tangent to (x-1) ^ 2 + (Y-1) ^ 2 = 1, and intersects with X-axis a, Y-axis B be quick!

Since the image of the straight line y = - x is located in the second and fourth quadrants, then the center of the circle is in the first and third quadrants. Suppose the center of the circle is m (a, a), and the radius of the circle is r = | a |, then the distance from the center of the circle m to the straight line x + y = 0 is:
D = | 2A | / √ 2 = 2, a = ± √ 2
Then the equation of the circle is: (x - √ 2) & # 178; + (Y - √ 2) & # 178; = 2 or (x + √ 2) & # 178; + (y + √ 2) & # 178; = 2
【2】
If a straight line is tangent to a circle, then the distance from the center of the circle to the straight line is equal to the radius of the circle
|n＋m－mn|/√(n²＋m²)=1----------------------------------------------(1)
Namely:
In addition, a (m, 0), B (0, n)
Then: S = (1 / 2) | Mn|
From (1), it is concluded that:
(n＋m－mn)²=m²＋n²
m²＋n²＋m²n²＋2mn－2m²n－2mn²=m²＋n²
m²n²＋2mn－2mn(m＋n)=0
mn＋2－2(m＋n)=0
2(m＋n)=mn＋2
Because m + n ≥ 2 √ (MN)
Then: Mn + 2 ≥ 4 √ (MN)
Let: √ (MN) = t, then: T & # 178; - 4T + 2 ≥ 0
2－√2≤t≤2＋√2
That is: the maximum value of √ (MN) is 2 ＋ 2, and the minimum value is 2 - √ 2
Then the maximum value of S is (1 / 2) (2 ＋ 2) & # 178; = 3 ＋ 2 √ 2, and the minimum value is (1 / 2) (2 － 2) & # 178; = 3-2 √ 2

From the first place to the second place, the car runs for 5 hours and the bus runs for 6 hours
The speed of a bus is about ()% of that of a car

83.33

Find the value of real number x, y, so that (Y-1) ^ 2 + (x + Y-3) ^ 2 + (2x + y-6) ^ 2 reaches the minimum, do not let a = Y-1. B = x + Y-3. C = 2x + y-6

In this problem, we first judge that (Y-1) ^ 2 + (x + Y-3) ^ 2 + (2x + y-6) ^ 2 ≥ 0, so we first try to see if we can satisfy that the minimum value is 0, (Y-1) ^ 2 + (x + Y-3) ^ 2 + (2X + y-6) ^ 2 = 0, then Y-1 = 0, x + Y-3 = 0 and 2x + y-6 = 0. Obviously, if there is no solution, then the minimum value is not 0

Party A and Party B paid the same amount of money to buy a box of apples. As a result, Party A took 6 kg and Party B took 14 kg. Party B had to give Party A 12 yuan. How much is the apple per kg?

This box of apples is 6 + 14 = 20kg
Party A and Party B should take the same Apple 20 / 2 = 10kg for the same amount of money
B took 14-10 = 4 kg, B to give a 12 yuan, then every kilogram of apple is 12 / 4 = 3 yuan