Five out of six times five out of nine plus five out of nine. One out of six has to have a picture!

Five out of six times five out of nine plus five out of nine. One out of six has to have a picture!

If the domain of definition of function y = f (x) = (X & # 178; - 2mx + 1) ^ 1 / 2 is x ∈ (- ∞, + ∞), find the value range of real number M

If the definition field of function y = f (x) = (X & # 178; - 2mx + 1) ^ 1 / 2 is x ∈ (- ∞, + ∞)
Then for any x ∈ R, there is always X & # 178; - 2mx + 1 ≥ 0
So Δ = 4m & # 178; - 4 ≤ 0
So - 1 ≤ m ≤ 1
That is, the value range of real number m is [- 1,1]

What numbers are 4 and 6 of 24

Common divisor

For the absolute value of formula x + 13, what is the minimum value of X? What is the minimum value?

/x/+13
/x/>=0
When x = 0
There is a minimum value = 13

What's 5 out of 12? What's 15 out of 8,

OK, is it degree?
5 / 12 * 60 = 25 points
8 / 15 * 60 = 32 points

1. If the value of the polynomial - X & sup2; + MXY + [n + 1] X & sup2; - 2XY + 5 with respect to the letter x.y is independent of the value of X, y, find the value of M + n

The original polynomial is reduced to NX & sup2; + (m-2) XY + 5, which has nothing to do with the value of X and y, so n = 0, M = 2, M + n = 2

There are two kinds of alcohol with 85% and 45% concentration. Now, to prepare 400 grams of alcohol with 60% concentration, how many grams should be extracted from each of these two kinds of alcohol

The best way to solve this kind of problem is to use the cross method. First, calculate the mass ratio of high and low concentration solutions
（60-45）/（85-60）=1/1
The mass ratio of 85% alcohol and 45% alcohol is 1 / 1. To prepare 400 grams of 60% alcohol, 200 grams of each alcohol should be extracted

13 out of 18 and 11 out of 12

13 out of 18 = 26 out of 36
11 out of 12 = 33 out of 36

3 / 10 - (3 / 4-7 / 10) 5 / 9 + (4 / 5 + 4 / 9)

3/10-（3/4-7/10）
= 3/10+7/10-3/4
=1-3/4
=1/4
5/9+（4/5+4/9）
=5/9+4/9+4/5
=1+4/5
=9/5

On the application of one variable quadratic
1. It is known that the sum of the coefficients and constant terms of the quadratic equation (KY + 1) (Y-K) = K-2 with respect to y is equal to 3. Find the value of K and the solution of the equation
2. If a + B + C = 0, then the quadratic equation AX square + BX + C = 0, there must be a solution____
If A-B + C = 0, then the quadratic equation AX square + BX + C = 0, there must be a solution____
If 4A + 2B + C = 0, then the quadratic equation AX + BX + C = 0 has a solution____
If 4a-2b + C = 0, then the quadratic equation AX square + BX + C = 0 has a solution____
Requirements. The second question directly to the answer on the line

1. The result is: KY square + (1-K Square) y-2k + 2 = 0
From the meaning of the title, K + 1-k square - 2K + 2 = 3, we can get k square + k = 0, k = 0 or - 1
When k = 0, y = - 2
When k = - 1, y = 1 + radical 5 or 1-radical 5
2、1；-1；2；-2