If the square of 3x - 2x + b-x-bx + 1 contains NO x term, B =?

If the square of 3x - 2x + b-x-bx + 1 contains NO x term, B =?

b -3
If a1 + A2 + a3 = 26, a4-a1 = 52, find an
A1 + A2 + a3 = 26, equivalent to A1 (1 + Q + Q * q) = 26, a4-a1 = 52, equivalent to A1 (Q * Q * Q-1) = A1 (Q-1) (Q * q + Q + 1) = 52, Q-1 = 2, so q = 3, bring in A1 = 2, an = 2 * 3n-1
a1=2 q=3
an=2*3^(n-1)
What is 1 + I divided by 3-4i?
1 + I divided by 3-4i
=(1+i) / (3-4i)
=(1+i)(3+4i) / (3+4i)(3-4i)
=(-1+7i) / 25
=(7i-1)/25
Multiply the numerator and denominator by 3 + 4I
= (1+i)(3+4i) / 25
= (-1+7i)/25
(7i-1)/13
If the square of 3x - 2x + 4 - (the square of X + BX-1) does not contain a linear term of X, then B = what
The square of 3x - 2x + 4 - (the square of X + BX-1)
=3x²-2x+4-x²-bx+1
=2x²-(2+b)x+5
∵ a term without x,
∴2+b=0
b=-2
Let {an} be an equal ratio sequence, A2 = 2, A5 = 1 / 4, find A1 * A2 + A2 * A3 + a3 * A4 + +an*a(n+1)=
Let the common ratio be Q
A5 = A2 * q ^ 3, q = 1 / 2
So A1 = 4, an = 4 * (1 / 2) ^ (n-1)
Let BN = an * a (n + 1) = 8 * (1 / 4) ^ (n-1)
So {BN} is an equal ratio sequence with the first term of 8 and the common ratio of 1 / 4
The first n terms of the original formula = BN and Sn = 8 * [1 - (1 / 4) ^ n] / (1-1 / 4) = 32 / 3 (1-1 / 4 ^ n)
If the complex Z ^ 4 = - 1, how much is Z? How to calculate?
Z ^ 4 = 1 * e ^ I π z = 1 * [cos ((π + 2K π) / 4) + isin ((π + 2K π) / 4)] (k = 0,1,2,3) Z has four values. In general, the solution of W ^ n = Z in the complex range is: z = R (COS θ + isin θ) (where R is the module of Z and θ is the radiation angle of Z) w = R ^ (1 / N) * [cos ((θ + 2K π) / N) + isin ((θ + 2K π) /
The second power of I = - 1
Z=i
z^-1
If there is no term containing x in the sum of the square of 3x - 2x + B and the square of X + BX-1, the value of B is obtained
3x & sup2; - 2x + B + X & sup2; + BX-1 = 4x & sup2; + (b-2) x + B-1, because there is no term containing x, B-2 = 0, B = 2
If a1 + A4 = 18, A2 + a3 = 12, then the sum of the first eight terms of the sequence is ()
A. 513B. 512C. 510D. 2258
Let the first term of the equal ratio sequence be A1, the common ratio be q ∵ a1 + A4 = 18, A2 + a3 = 12 ∵ A1 (1 + Q3) = 18a1q (1 + Q) = 12. By dividing the two formulas, 2q2-5q + 2 = 0 can be obtained by taking the common ratio Q as an integer, q = 2, A1 = 2 can be obtained by substituting it into the sum formula of the equal ratio sequence, S8 = 2 (1 − 28) 1 − 2 = 510
When z = - 1-i2, the value of Z100 + Z50 + 1 is equal to______ .
∵ z = - 1-i2 = 22-22i ∵ Z2 = 12-2 × 22 × 22I + (22I) 2 = - I, we can get Z4 = - 1. According to the meaning of plural power, we can get Z100 = (Z4) 25 = - 1, Z50 = (Z4) 12 · Z2 = - I ∵ Z100 + Z50 + 1 = - 1-I + 1 = - I, so the answer: - I
Let {x = Ӝ - x} be a set=
quickly
A={x|x²-4x≤0,x∈R}
x²-4x≤0
X(X-4)≤0
0≤X≤4
So a = {x | 0 ≤ x ≤ 4}
B={y|y=-x²}
Because x ∈ R, X & # 178; ≥ 0, - X & # 178; ≤ 0
B={y|y≤0}
A∩B={0}
CR(A∩B)=（-∞,0）∪（0,+∞)