In the expansion of (x − 1x) 9, the coefficient of X3 is______ ．

In the expansion of (x − 1x) 9, the coefficient of X3 is______ ．

TR + 1 = CR9 (x) 9 − R (− LX) r  c9r (x) 9-2r (- 1) r, let 9-2r = 3 {r = 3, the coefficient of X3 is T4 ′ = C93 (- 1) 3 = - 84

In the expansion of binomial (x-1x) 9, the coefficient of X3 is______ ．

Because the general formula of binomial (x-1x) 9 is tr + 1 = CR9 · x9-r · (- 1) r · X-R = (− 1) r · C & nbsp; R9 · x9-2r, let 9-2r = 3, the solution is r = 3, and the coefficient of X3 in the expansion is (− 1) 3 · C & nbsp; 39 = - 84, so the answer is - 84

In the expansion of (1-3x) ^ n, the sum of the coefficients is equal to 64?

（1-3x）^n
The sum of each system is 64. Take x = 1 and the above formula is the sum of each coefficient
（1-3）^n＝64
The solution is n = 6
(1-3x) ^ 4 contains x ＾ 2 items
C（6,2）*（－3X）＾2.
Coefficient = 9 * C (6,2) = 135

In the expansion of binomial (1-3x) n, if the sum of the coefficients of all terms is equal to 64, then n=______ In this expansion, the coefficient with x2 term is______ ．

In (1-3x) n, let x = 1, then the sum of coefficients of all terms is (- 2) n, ∧ - 2) n = 64. The general term of the expansion of n = 6 ∧ (1-3x) n = (1-3x) 6 is tr + 1 = c6r (- 3x) r = (- 3) rc6rxr. Let r = 2, then the coefficient of the expansion with x2 term is 9c62 = 135, so the answer is 6135

If the sum of the coefficients in the (3x-1x) n expansion is 32, then the coefficient of the term with X3 in the expansion is ()
A. -5B. 5C. -405D. 405

Let x = 1 get the sum of the coefficients of the expansion as 2n ﹥ 2n = 32, the solution is n = 5 ﹥ 3x-1x) n = (3x-1x) 5, the general term of the expansion is tr + 1 = (- 1) r35-rc51x5-2r, let 5-2r = 3 get r = 1, so the coefficient of the term with X3 in the expansion is - 34c51 = - 405, so C is selected

In the expansion of (x ^ 2 + 3x + 2) ^ 4, there are coefficients of x ^ 5
The correct answer is 180

In the expansion of (X & # 178; + 3x + 2) &# 8308;, the coefficients of the term x & # 8309; are included
The original formula = (x + 1) & # - 8308; (x + 2) & # - 8308; = (X & # - 8308; + 4x & # - 179; + 6x & # - 178; + 4x + 1) (X & # - 8308; + 8x & # - 179; + 24x & # - 178; + 32x + 16)
So the coefficient of X & = 1 × 32 + 4 × 24 + 6 × 8 + 4 × 1 = 32 + 96 + 48 + 4 = 180

In the expansion of (2x-3x Λ 3) Λ10, the coefficient of X Λ 16 is?

If in the product of (2x square - 3x + a) (3x square + x-1), the coefficient of the square term of X is 16, then a =?

(2x square - 3x + a) (3x square + x-1)
=The fourth power of 6x + 2x & # 179; - 2x & # 178; - 9x & # 179; - 3x & # 178; + 3x + 3ax & # 178; + ax-a
=The fourth power of 6x - 7x & # 179; - (2 + 3-3a) x & # 178; + 3x + ax-a
-(2+3-3a)=16
3a=21
a=7.

10.5 (x-4) = 2.5x, thank you

21（x-4）=5x;
21x-5x=84;
16x=81;
x=81/16;

0.4x-0.3-x = 0.4 times (0.5x-3) to find x
Please leave the calculation steps. Both - 21 / 11 and 27 / 11 are wrong.

10x/4-1.2/4-4x/4=4/5*(x/2-6/2)
(50x-6-20x)/20=(8x-48)/20
therefore
30x-6=8x-48
Solution x = - 21 / 11