Factorization a ^ 2 + B ^ 2-2ab-a + B

Factorization a ^ 2 + B ^ 2-2ab-a + B

（a-b)²-(a-b)
(a-b)(a-b-1)

A={x|x=1/9(2k+1),k∈Z}
B={x|x=4k/9±1/9,k∈Z}
What is their relationship?
A={x|-3≤x≤4}
B{x|x≥m}
And a is contained in B
Then the value range of real number m is____ .

First: a = B
Second: M

A factorization problem
3^2000+6*3^1999-3^2001
Need process!

3^2000+6*3^1999-3^2001
=3*3^1999+6*3^1999-3^2*3^1999
=3^1999*(3+6-3^2)
=3^1999*0
=0

By using trigonometric function lines in the unit circle, we can find the set of X satisfying cosx-1 / 2
There is no picture

That's zero to one hundred and two, two hundred and four to three hundred and six

A factorization problem
x^2-x+(1/4)

x^2-x+(1/4)
=x^2-2*x*(1/2)+(1/2)^2
=(x-1/2)^2

Find SiNx > 1 / 2 and cosx

The graph of SiNx & gt; 1 / 2 corresponding terminal edge in the coordinate system is as follows: (excluding boundary)
The graph of the terminal edge corresponding to cosx & lt; 1 / 2 in the coordinate system is as follows: (excluding the boundary)
Take the intersection above,
The set of angles of SiNx & gt; 1 / 2 and cosx & lt; 1 / 2 is {x | 2K π + π / 3 & lt; X & lt; 2K π + 5 π / 6, K ∈ Z}

How to use trigonometric function line to find cosx ≥ 1 / 2? Trigonometric function line

Draw trigonometric function lines
Where the cosine is on the x-axis
So it's positive on the right side of the y-axis
Cosx = 1 / 2 corresponds to 2K π ± π / 3
If it is more than 1 / 2, it is on the right side of 1 / 2
So 2K π - π / 3 ≤ x ≤ 2K π + π / 3

Using trigonometric function line, find the set of angles a satisfying the following conditions, sin = 1 / 2, sin is less than or equal to - 1 / 2

"Sin = 1 / 2, sin less than or equal to - 1 / 2" please explain

The limit of [x * sin (3 / x)], X tends to be positive infinity

lim(x→∞)x*sin(3/x)
t=1/x
=lim(t→0)sin(3t)/t=lim(t→0)3sin(3t)/(3t)=3

Find the limit of x [(sin (1 / x)] (x tends to infinity)

X tends to infinity and 1 / X tends to zero
Sin (1 / x) tends to 1 / X
So the limit is 1