·(1.3:2 plus 2.4:3 plus 3.5:4 plus 4.6:5) divided by (3.3:1 plus 5.4:2 plus 7.5:3 plus 9.6:4) is equal to To have a process, this is a simple operation

·(1.3:2 plus 2.4:3 plus 3.5:4 plus 4.6:5) divided by (3.3:1 plus 5.4:2 plus 7.5:3 plus 9.6:4) is equal to To have a process, this is a simple operation

·(1 and 3 / 2 + 2 and 4 / 3 + 3 / 4 + 4 / 5 + 5 / 6) = (1 + 2 + 3 + 4 + 2 / 3 + 3 / 4 + 4 / 5 + 5 / 6) / (3 + 5 + 7 + 1 / 3 + 2 / 4 + 3 / 5 + 4 / 6) = (10 + (2 / 3 + 5 / 6) + 3 / 4 + 4 / 5) / (24 + (1 / 3 + 4 / 6) + (2 / 4 + 3 / 5)) = (10 + 6

13-8 = () - 6 = 12 - () 15-7 = () + 3 = 12 - () 12-6 = () - 9 = () - () what is the law

That's how it should be calculated
13-8=（11）-6=12-（7） 15-7=（5）+3=12-（4） 12-6=（15）-9=（8）-（2）
The rule is that the sum of them is the same answer!

7.4(x+5)-6.5x=122.3

Solution
7.4x+37-6.5x=122.3
0.9x=122.3-37
0.9x=85.3
x=85.3÷0.9
x=853/9

The surface area of a cylinder is 628 square centimeters. The bottom radius of the cylinder is 1 / 3 of the height. Calculate the side area of the cylinder

If the bottom radius is a, the height is 3a
Surface area = 2 π A & sup2; + 2 π a × 3A
=6.28a²+6,28×3a²
=（6.28×4）a²
=25.12a²
=628
Then a & sup2; = 25. A = 5
Side area = 2 π a × 3A = 6.28 × 3 × 25 = 471

It is known that a * 10 / 3 = 11 / 12 * b = 15 / 15 * C, and A.B.C is not equal to 0. The three numbers of A.B.C are arranged from small to large, and why?

a×10/3=b×11/12=c×15/15
a×10/3=b×11/12
a=b×11/12×3/10
a=b×11/40
=b-b×29/40

When LG (︱ x + 3 ︱ X-7 ︱ a, a, a) is a value, the solution set is r

a=lg10^a
So | x + 3 | + | X-7 | > 10 ^ a
Because | x + 3 | + | X-7 | = | x + 3 | + | 7-x | > = | x + 3 + 7-x | = 10
Therefore, if | x + 3 | + | X-7 | > 10 ^ A is established, then 10 > 10 ^ a
So a

One fifth of 40 is equivalent to a number multiplied by a fraction of 60. The product is either greater than this number or less than this number () nine tenths multiplied by nine tenths is greater than nine tenths () the number of class one is equal to eleven tenths of class two, and the number of class one is one tenth more than that of class two () increase the 30m by one fifth and then decrease by one fifth, Because 15 / 7 times 7 / 15 = 1, so 15 / 7 and 7 / 15 are reciprocal () 1 △ a = B (A and B are not 0), a and B must be reciprocal

One fifth of 40 is equal to two fifths of 60
If a number is multiplied by a fraction, the product is either greater than the number or less than the number (×)
Nine tenths times nine tenths is greater than nine tenths (×)
The number of people in class one is equal to 11 / 10 of that in class two. The number of people in class one is one tenth more than that in class two (×)
First increase 30 m by one fifth and then decrease it by one fifth, or 30 m (×)
The meanings of 2 / 9 times 8 and 8 times 2 / 9 are different, but the calculation method and result are the same (√)
Because 15 / 7 times 7 / 15 = 1, so 15 / 7 and 7 / 15 are reciprocal (×)
1 △ a = B (A and B are not 0), a and B must be reciprocal to each other

A new operation is defined for rational number x.y: X * y = ax + by + 5, where A.B is a constant; given 1 * 2 = 9, (- 3) * 3 = 6, then the value of 2 * 7 is
A new operation is defined for rational number x.y: X * y = ax + by + 5, where A.B is a constant; given 1 * 2 = 9, (- 3) * 3 = 6, then the value of 2 * 7 is________ .

a+2b+5=9 3a+6b=12
-3a+3b+5=6 -3a+3b=1
9b=13 b=13/9;a=10/9
2*7=20/9+91/9=111/9=37/3

19 minus 1

It's 19-1.2x = 7
19-1.2x=7
1.2x=19-7
1.2x=12
x=10

Let x, y ∈ R + and XY - (x + y) = 1, then the minimum value of X + y is______ ．

∵ XY - (x + y) = 1, ∵ xy = (x + y) + 1 ∵ XY ≤ (x + Y2) 2, ∵ (x + y) + 1 ≤ (x + Y2) 2 = 14 (x + y) 2, we can get (x + y) 2-4 (x + y) - 4 ≥ 0, let t = x + y, we can get t2-4t-4 ≥ 0, the solution is t ≥ 2 + 22 (rounding) ∵ x + y ≥ 2 + 22, we can get the minimum value of X + y is 2 + 22, so the answer is: 2 + 22