A = (2,1), B = (- 1,0), then 3a-2b = how to calculate?

A = (2,1), B = (- 1,0), then 3a-2b = how to calculate?

It's a vector, right
3a=(6,3)
2b=(-2,0)
3a-2b=(6+2,3+0)=(8,3)

Calculation: 2A ^ 2B (3a-b + 0.5)

2a^2b(3a-b+0.5)=6a³b-2a²b²+a²b

(3a-2b)(9a+6b)

If 3a-2b = 1, then 6b-9a + 3 =?, fast

3a-2b=1
2b-3a=-1
6b-9a=-3
6b-9a-3=0

If 3A + 2B = 23, then 9A + 6B=

b-(a-b)=2b-a
3(3a+2b)=3*23=69

In the same coordinate system, if there is an ordinal pair (3a + 1, b-2) with the same position as (- 5,1), then a =? B =?
emergency

3a+1=-5 3a=-6 a=-2
b-2=1 b=3

3A-3[2B-8+(3A-2B-1)-A]+1
Sort out the brackets

Given / A / = 5 / B / = 9, find the value of 3A + 2B

Sanyuan Yishi: a + 2B + C = 9 - 3A + 3B + C = 6, B + C = 2

It is known that a and B satisfy B = √ A-2 + √ (6-3a) + 5, and the reduction calculation is: √ a · √ A / b ÷ 2B / a × √ 2B / A
Reduction calculation: √ a × √ A / b ÷ 2B / a × √ 2B / A

If √ A-2 + √ (6-3a) is significant, then A-2 ≥ 0, that is, a ≥ 2
6-3a ≥ 0, i.e. a ≤ 2
So: a = 2, B = 5
√a×√a/b÷√2b/a×√2b/a
=a/√b÷2b/a
=a²/(2b√b)
=4/[10√5]
=2√5/[25]