The problem of sequence limit: "LIM (n →∞) an = a" is a () (A) Sufficient and nonessential condition (b) necessary and nonessential condition (c) necessary condition (d) nonessential and nonessential condition,

The problem of sequence limit: "LIM (n →∞) an = a" is a () (A) Sufficient and nonessential condition (b) necessary and nonessential condition (c) necessary condition (d) nonessential and nonessential condition,

D. If the sequence is like this: a + 1, a + 2, a + 1 / 3, a + 1 / 4.. a + 1 / n.. Then it still satisfies LIM (n →∞) an = a, but not decreasing.. then if the sequence is: {a + 1 + 1 / n}, decreasing but not LIM (n →∞) an = a

Find the difference between 2A & # 178; b-4c + 1 and - 3A & # 178; B + 4d-1. If a and B are reciprocal and C and D are opposite, simplify the difference

(2a²b-4c+1)-(-3a²b+4d-1)
=2a²b-4c+1+3a²b-4d+1
=5a²b-4(c+d)+2
If a and B are reciprocal, C and D are opposite
Then AB = 1, C + D = 0
5a²b-4(c+d)+2=5(ab)b-4(c+d)+2=5b+2

Let Tan (a + b) = 2tan a prove 3sinb = sin (2a + b)

To prove 3sinb = sin (2a + b)
That is to say, 3sin (a + B-A) = sin (a + B + a)
That is to say, 3sin (a + b) cosa-3cos (a + b) Sina = sin (a + b) cosa + cos (a + b) Sina
That is to say, 2Sin (a + b) cosa = 4cos (a + b) Sina
That is to say, Tan (a + b) = 2tana
From the known conditions, it is proved

Which algorithm structure should be used in the algorithm of finding the approximate root of equation x2-2 = 0 by dichotomy ()
A. Sequential structure B. conditional structure C. cyclic structure D. all of the above

Any algorithm has a sequential structure, and the cyclic structure must contain a conditional structure. The dichotomy method uses the cyclic structure to find the approximate root of the equation x2-2 = 0. The algorithm uses sequential structure, conditional structure and cyclic structure, so D is selected

The solution of the basic solution system of linear algebra
Let one of the free unknowns be 1, and the rest be zero
The basic solution system of homogeneous equations is not well understood

Take homogeneous equations as an example: if it is a third-order matrix R (a) = 1, there is only one equation left after matrix transformation. At this time, you can set X3 as 1, X2 as 0, and get x1. Then you can set X3 as 0, X2 as 1, and get x1. You may wonder why you want to set it like this, and why you should set it like this. The reason is very simple, because as long as (0,1) and (1

Find the length of the chord ab of the line L: 3x-y-6 = 0 cut by the circle C: x2 + y2-2x-4y = 0

Change x2 + y2-2x-4y = 0 to the square of (x-1) + (Y-2) = 5, then the distance from the center of the circle (1,2) to the straight line L is equal to the radius, and then it can be calculated by using the distance formula from the point to the straight line

Given that x is an integer, and X + 2 / 3 + 3-2 / X & # 178; - 2x + 18 / 9 is an integer, find X that meets the condition

If the condition is satisfied that the denominator cannot be 0, then x + 3 ≠ 0, 3-x ≠ 0, X & # 178; - 9 ≠ 0. If x ≠ ± 3 is obtained by solving the three equations, the condition is satisfied and can be expressed as a set

It is known that a = ax ^ 2 + X-1, B = 3x ^ 2-2x + 1 (a is a constant). (1) if the sum of a and B does not contain the square term of X, then a=_______ （2

A+B=ax^2+x-1+3x^2-2x+1=(a+3)x^2-x-2
Without square term
a+3=0
a=-3

Given the function f (x) = 2 to the power of X, X ∈ R, if the inequality f (x) + F (x) - M > 0 holds on R, the range of M is obtained

Because x ∈ R, the range of F (x): (0, + infinity)
Let k = f (x) (k > 0), the square of F (x) + F (x) - M > 0, that is k ^ 2 + k > M
Let g (k) = k ^ 2 + K, so M0
So m ≤ 0

Given that the equation of circle C is the square of (x-1) + the square of y = 9, the tangent equation of circle m (- 2,4) is solved

The center of circle C: (X -- 1) ^ 2 + y ^ 2 = 9 is C (1,0), and the radius is 3,
Let the tangent equation of circle C passing through point m (- - 2,4) be:
Y -- 4 = K (x + 2), (k is the slope of the straight line),
That is: KX -- y + 2K + 4 = 0,
Because the distance from the tangent to the center of a circle is equal to the radius of the circle,
So ik-0 + 2K + 4I / radical [K ^ 2 + (-- 1) ^ 2] = 3
That is: i3k + 4I = 3 radical (k ^ 2 + 1)
Square of both sides: 9K ^ 2 + 24K + 16 = 9K ^ 2 + 9
24k=--7
k=--7/24
So the tangent equation of the circle is:
y--4=--7/24(x+2).
That is: 7x + 24y = 82
The other line x = -- 2 is also tangent to the circle C,
Because the distance from it to the center of the circle C (1,0) is also equal to the radius of the circle C 3