When a = 1 / 6, B = 1 / 2, 3A + 2B = ()

When a = 1 / 6, B = 1 / 2, 3A + 2B = ()

3a+2b=3*1/6+2*1/2=1/2+1=3/2

Given 3 Λ a = 2, 3 Λ B = 6, find the value of 3 Λ 3a-2b

The original formula = (2 ＾ 3) / (6 ＾ 2) = 2 / 9

a=-1/6 b=3/2 3a+b/3a-b-3a-b/3a+b/[1/(1-b/3a)]/(1-2b/3a+b)=?

3a+b/3a-b-3a-b/3a+b/[1/(1-b/3a)]/(1-2b/3a+b)
=(3a-3a)+(b/3a-b/3a)-b+b*(1-b/3a)/(1-2b/3a+b)
=-b+(1-1/3a)/(1/b-2/3a+1)
=-3/2+(1+2)/(2/3+4+1)
=-3/2+9/17
=-33/34

If / A / = 3, / B / = 2, and a > b, find the value of 3a-2b!

The answer is: 5 or 13
The value of a must be 3, B - 2 and 2

3a^3+2b^3>=3a^2b+2ab^2

2ab^2(a^2+2a^2b+b^3+3a)

2a³b²+4a^(2b+1)b²+2ab^5+6a²b²

The positions of rational numbers a, B and C on the number axis are shown in the figure, and | a | = | B | (1) degenerate and evaluate: | a | - | a + B | - | C-A | + | C-B | + | AC | - | - 2b | (2) if | a | = 3, | B | = 2 and | ab | = AB, calculate the value of 3a-2b

(1) The known number axis is: a = -b, that is, a + B = 0, a + B = 0, and the known number axis is: a = -b, that is, a + B = 0, and the known number axis is: a = -b, that is, a + B = 0, that is, a + B = a + B |||a |a ||b | ||||||||||||||||||\\\\\\\\\\124whena = 3, B = 2 or a = - 3, B = - 2, 3a-2b = 3 × 3-2 × 2 = 5 when a = 3, B = 2, 3a-2b = 3 × (- 3) - 2 × (- 2) = - 5 when a = - 3, B = - 2

Find the value of a = 2, B = 1.5, 3a-b =? A + B =? 40 divided by a =? B2 =? (6a) divided by (2b) +? (2a) 3 =? 8 (a-b) =? 10-a-b

3a-b==6-1.5=4.5
a+b==2+1.5=3.5
40 divided by a = 40 △ 2 = 20
b²=1.5²=2.25
(6a) divided by (2b) = 12 △ 3 = 4
（2a）³==4³=64
8（a-b）=8×(2-1.5)=4
10-a-b=10-2-1.5=6.5

a:b=7:8,2b:a=( ):( ),2b:3a=( ):( )

Let a * b = A2 + 2B, then 10 * 6 and 5 * (2 * 8) let a * b = 3a-b × 1 / 2, then (25 * 12) * (10 * 5)

First question 112 second question 65 third question 193.25