Given a = 3x2-2x + 1, B = 2x3-x2 + 3x-7, find 3a-2b

Given a = 3x2-2x + 1, B = 2x3-x2 + 3x-7, find 3a-2b

3A-2B
=3*(3x²-2x+1)-2*(2x³-x²+3x-7)
=9x²-6x+3-4x³+2x²-6x+14
=-4x³+11x²-12x+14
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2X2—2X—3X2—15X+18>1+4X—X2
This is an inequality
X = unknown
X is followed by times
Please have a look

2x^2-2x-3x^2-15x+18>1+4x-x^2
-x^2-17x+18>-x^2+4x+1
-21x>-17
x

The minimum value of the function f (x) = 3x4-2x3-3x2 is

Let f ′ (x) = 12x & # 179; - 6x & # 178; - 6x = 12x (x + 1 / 1) (x-1) = 0, then x = 0, x = - 1 / 2 or x = 1. When x ＜ - 1 / 2 or 0 ＜ x ＜ 1, f ′ (x) ＜ 0, the function decreases monotonically; when x ＞ 1 or - 1 / 2 ＜ x ＜ 0, f ′ (x) ＞ 0, the function increases monotonically. Therefore, the function has a maximum value f (0) = 0, but no maximum value

Given that {an} is an increasing sequence and an = n ^ 2 + λ n holds for any n ∈ n +, then the value range of real number λ is ()
How to do this problem with quadratic function image?
-λ/2

A:
Let f (n) = an = n & # 178; + λ n
An is a monotone increasing sequence
So: F (n) is a monotone increasing function
Because: F (n) is a parabola with an opening upward
Axis of symmetry n = - λ / 2-3
The midpoint of n = 1 and N = 2 is 3 / 2

What is the remainder of 586 3-phase multipliers divided by 5

The remaining 3:
⑴586×3=1758
⑵1758÷5=351.3

Use a simple method to calculate the following questions. 5.6 ﹣ 3.59.6 ﹣ 0.8 ﹣ 0.44.2 × 99 + 4.217.8 ﹣ 1.78 × 4

①5.6÷3.5=5.6÷0.7÷5=8÷5=1.6；②9.6÷0.8÷0.4=9.6÷（0.8×0.4）=9.6÷0.32=30；③4.2×99+4.2=4.2×（99+1）=4.2×100=420；④17.8÷（1.78×4）=17.8÷1.78÷4=10÷4=2.5．

Parametric equation of conic
I like to use parametric equation to solve this kind of problem when there is a straight line but the origin intersects the ellipse at two points a and B. But when using parametric equation to represent two points a and B, only one of them will be represented, and the other will not. If not, please explain the reason

Directly let a (acost, bsint), B (acosu, bsinu)

If mm-1; X
If mm-1; X

Because M

6G = fraction of kg

6g=6\1000kg=3\500kg

In the sequence {an}, A3 = 1, a1 + A2 + +an=an+1（n=1，2，3…） (I) find A1, A2; (II) find the first n terms and Sn of sequence {an};

(I) ∵ A1 = A2, a1 + A2 = A3, ∵ 2A1 = A3 = 1, ∵ A1 = 12, A2 = 12. (II) ∵ Sn = an + 1 = Sn + 1-sn, ∵ 2Sn = Sn + 1, Sn + 1sn = 2, ∵ Sn} is an equal ratio sequence with S1 = A1 = 12 and common ratio of 2