1, the definition field of function f (x) = log2 (4x minus 3) minus log2 (2 minus x) is? 2, let vector a = (2,4), B = (1,1), if B is vertical (a + m times b), then real number M =?

1, the definition field of function f (x) = log2 (4x minus 3) minus log2 (2 minus x) is? 2, let vector a = (2,4), B = (1,1), if B is vertical (a + m times b), then real number M =?

Greater than or equal to 3 / 4 and less than or equal to 2

What is the solution of the equation sin (x / 2) = 1 / 3 on (pie, 2 pie)
I don't think there is any solution

π/2

It is known that the equation SiN x + cos x = k has two solutions on the 0 ≤ x ≤ pie, and the value range of K is obtained

SiN x + cos x = radical 2Sin (x + π / 4)
0 ≤ x ≤ p
So π / 4 ≤ x + π / 4 ≤ π 5 / 4
So the radical 2 ≤ 2Sin (x + π / 4) ≤ 2
So root 2 ≤ K < 2

Given two natural numbers a and B (a > b), if the remainder of a and B divided by 7 is 3 and 6 respectively, find the square of a + B A-B A-B the square of B the square of a + B
Given two natural numbers a and B (a > b), if the remainder of a and B divided by 7 is 3 and 6 respectively, find the square of a + B A-B A-B the square of A-B the square of a + B and find the remainder of each divided by 7

Let a / 7 = x-3, B / 7 = y.6
Then a = 7x + 3, B = 7Y + 6
A + B = 7 (x + y + 1) + 2 is the remainder 2
A-B = 7 (x-y-1) + 4 is the remainder 4
A ^ 2-B ^ 2 = 7 (x ^ 2-y ^ 2 + 6x-12y-4) + 1 is the remainder 1
A ^ 2 + B ^ 2 = 7 (x ^ 2 + y ^ 2 + 6x + 12Y + 6) + 3 is the remaining 3

It is known that the polar coordinate equation of the parabola is ρ = 4 / 1-cos θ, and the polar coordinate equation of the Quasilinear of the parabola is obtained

【1】 My method is as follows: first, it is transformed into a general equation to solve, and then restored. [2] the general equation of the parabola is Y & sup2; = 8 (x + 2). It is easy to know that the Quasilinear equation is x = - 4. The polar coordinate equation of the Quasilinear is ρ cos θ = - 4

(a+b）^2-16a^2
(x^2+4)^2-4x^2
m^4-2m^2n^2+n^4
(x^2-x)^2+1/2（x^2\x)+1/16
4m^2-n^2-6n-9
2a-4b+a^2-4b^2
(3a+b)^2-12ab
(m-1)(m-3)+1
x^2(x^2-2)+1
-x^2-12xy-20y^2
3x^2-11xy+10y^2
2(a-1)-11(a-1)+12
9a^4-7a^2+1
a^4+a^2b^2+b^4
The fourth question is (x ^ 2-x) ^ 2 + 1 / 2 (x ^ 2-x) + 1 / 16

Given that the line y = - radical 3 / 3x + radical 3 intersects the x-axis at point a, intersects the y-axis at point C, and points B on the coordinate axis, the triangle ABC is isosceles triangle, and the base angle is equal to 30 degrees, then the coordinate of point B is____
(there is no picture for this question. You should draw your own picture to make it.)

If (x-y-3) &# 178; and | 4x-3x + 11 | are opposite numbers, then the value of X and Y is the solution step

(x-y-3) &# 178; and | 4x-3x + 11 | are opposite to each other, so (x-y-3) &# 178; and | 4x-3x + 11 | are both x-y-3 = 04x-3y + 11 = 0, and the solution is: x = - 20Y = - 23_ If you don't understand, you can continue to ask this question, and wait online at any time

When k =, the parabola y = x ^ 2 + 4x + KX has only one intersection with the coordinate axis
The title is wrong. sorry.
When k = k, the parabola y = x ^ 2 + 4 + KX has only one intersection with the coordinate axis

y=x^2+(k+4)x
There is always an intersection between a quadratic function and the y-axis
So there's no intersection with the x-axis or at the origin
If there is no intersection with the X axis
So the discriminant is less than 0
k^2-16

The formula of equivalent relation in encounter problem and pursuit problem

Distance = velocity and X encounter time
Pursuit distance = speed difference × pursuit time. Great! Good luck