In the expansion of binomial (2x + 6) ^ 6, what is the coefficient with x ^ 4 term

In the expansion of binomial (2x + 6) ^ 6, what is the coefficient with x ^ 4 term

(16x ^ 4 + 192x ^ 3 + 864x ^ 2 + 1728x + 1296) x (4x ^ 2 + 24x + 36) take out the x ^ 4 term in the product and calculate its value, which is equal to
576x ^ 4 + 4608x ^ 4 + 3456x ^ 4 = 8640x ^ 4, so the coefficient of x ^ 4 is 8640

In the n-th expansion of binomial (square of X + 1 / x), if the binomial coefficients of the fourth and seventh terms are equal, the expansion is obtained
Finding the constant term in the expansion

The key is to understand, don't be afraid of trouble, (a + b) ^ n = [cn (n is the subscript) 0 (0 is the superscript)] XA ^ nxb ^ 0 (x is the multiplication sign for convenience) + [cn (n is the subscript) 1 (1 is the superscript)] XA ^ n-1xb + +[cn (n is subscript) n (n is superscript)]. This problem is CN (n is subscript) 3 (6 is superscript)] = CN (n is

High school mathematics: in the expansion of binomial (AX-1 / x) ^ 8, the constant term is 70, then the real number a =?
I'm rather stupid. I didn't study this kind of problem well, alas
Sorry, wrong number, is: in the expansion of binomial (AX-1 / radical x) ^ 8, the coefficient of x ^ 2 is 70, then the real number a =?

T(r+1)=C(n,r)a^(n-r)b^r
=>
T(r+1)=C(8,r)(ax)^(8-r)(-x^(-1/2)^r=[C(8,r)(a^(8)(-1)^r]x^(8-r-r/2)
=>8-3r/2=0
=>r=16/3
There is a problem with the question? R must be an integer!

In the expansion of binomial (AX-1 / x) ^ 8, if the constant term is 70, then the real number a =?
I'm rather stupid. I didn't study this kind of problem well, alas

Is a binomial [ax - (x under 1 / radical)] ^ 8, then there is no constant term
Is a binomial [(AX-1) / x under the radical] ^ 8, then it becomes
[a√x-(1/√x）]^8
In the expansion, the constant term is C (subscript 8, superscript 4) * (a √ x) ^ 4 * (1 / √ x) ^ 4 = 70A ^ 4
In the expansion of binomial (AX-1 / x) ^ 8, if the constant term is 70, then
70a^4=70
a=±1

Why is the function y = x / TaNx, x = k π + π / 2 (k = 1,2,3. - 1, - 2, - 3.) a separable breakpoint

In k = - 3, - 2, - 1,, 1,2,3,, the left and right limits are the same, they are all zero. TaNx tends to positive and negative infinity respectively, and 1 / TaNx tends to zero from positive and negative respectively

Find the function f (x) = x (x-1) (X-2) The derivative of (X-100) at x = 0

f(x)=x^101+(a1)x^100+(a2)x^99+…… +(a99)x^2+(a100)x
So the derivative of F (x) at x = 0 is A100
That is, the coefficient of X in F (x)
And f (x) = x (x-1) (X-2) (x-100)
So the coefficient is (- 1) × (- 2) × (- 3) × (- 4) × Factorial of × (- 99) × (- 100) = 100
That is 100!

The continuity and differentiability of function f (x) at point x = 0 are discussed

Given that the function f (x) is differentiable on [0,1], f (x) > 0, f (0) = 1, and satisfies the equation f (x) - 1 / (x-1) ∫ (1, x) TF (T) DT = 0 in [0,1], find the function f (x)

The results show that: (x-1) f (x) -∫ (1, x) TF (T) DT = 0
F (x) + (x-1) f '(x) - XF (x) = 0, that is: F' (x) - f (x) = 0, the solution of the equation is: F (x) = CE ^ X
From F (0) = 1, C = 1, so f (x) = e ^ X

F (x) = x & # 178; + ∫ (1,0) XF (T) DT + ∫ (2,0) f (T) DT find function f (x)

In this paper, we will be the (1,0) f (1,0) and (2,0) f (T) let (1,0) f (1,0) f (T) let (1,0) f (1,0) and (1,0) f (T) DT = a (2,0) and (2,0) f (t) in the (1,0) f (1,0) f (1,0) f (T) in the (1,0) and (1,0) f (1,0) f (1,0) f (t (T) in the (T) in the following (x) \\\8747 (1,1,0) in this (1,1,0) in the following (1,1,0) we (1,1,0) in this (1,1,1,0) in this (1,1,0) in this (1,1,1,1,0) in this (1,1,0) and on both sides of + ax + B

The discontinuous point of the function f (x) = 1 / 1-e Λ x is
Please write down the steps to solve the problem

The definition field of ∵ f (x) is 1-e ^ x ≠ 0, that is, X ≠ 0
X = 0 is the discontinuity of F (x)
When x → 0 +, limf (x) = - ∞
When x → 0 -, limf (x) = + ∞
X = 0 is an infinite discontinuity in the second kind of discontinuities of F (x)