One or two digits, the sum of the numbers is 9. If the transposition of one digit and ten digit is multiplied by the original number, the result is 1458. If the ten digit is x, the equation is? 1. The single digit is 9-x 2. The original number is 9-x + 10x (9-x) * 10 + X 4. Because the multiplication of two numbers is 1458 So (9-x + 10x) ((9-x) * 10 + x) = 1458 I don't understand the answer, please explain it in detail

The original number is 9-x + 10x, the ten digits are x, and each digit is 9-x. the number of ten digits should be x 10. For example, 84 = 8 * 10 + 4, so this number is 10x + 9-x3

For a two digit number, the ten digit number is x and the single digit number is y. transpose the ten digit number and the single digit number to get a new two digit number, and calculate the sum and difference between the new number and the original number

The original number is 10x + y 10Y + X
Then the sum is 10x + y + 10Y + x = 11 (x + y)
The difference is 10x + y-10y + x = 9 (X-Y)

For a two digit number, the ten digit number is x and the single digit number is y. if you swap the ten digit number with the single digit number, you will get a new two digit number. Please calculate the difference between the new number and the original number
This can be divided by 11. What's the difference?

The original number is 10x + y
The new number is 10Y + X
The sum of them is 10x + y + (10Y + x) 11x + 11y = 11 (x + y)
So their sum is a multiple of 11
The difference is 10x + Y - (10y-x) = 9 (X-Y)
So their difference is not divisible by 11

p1(x1,y1) l:f(x,y)=0 p2(x2,y2) f(x,y)+f(x1,y1)+f(x2,y2)=0
P1 (x1, Y1) is the point on the line L: F (x, y) = 0, P2 (X2, Y2) is the point outside the line L,
Then f (x, y) + F (x1, Y1) + F (X2, Y2) = 0?
Someone gave the answer
(because P1 is on a straight line, f (x1, Y1) = 0
Since P2 is not on the line, f (X2, Y2) is not zero
f(x,y）+ f(x1,y1) + f(x2,y2) ≠ f(x,y)
Because the slope of the line on the left is the same as that of the far line L
So parallel)
The slope of the equation is the same as that of F (x 2, Y 1) + F (Y 1) + Y 1, but why not

Because when f (x1, Y1) = 0
A straight line is f (x, y) = 0
The other line is f (x, y) + F (X2, Y2) = 0
Here P (X2, Y2) is just a point outside the line. Then f (X2, Y2) is a definite function value
Therefore, f (x, y) + F (X2, Y2) = 0 is equivalent to f (x, y) = 0 translating f (X2, Y2) distance units
Parallel, of course
In addition, you can get the slope from the derivative method to consider
The slope of the first line is [f (x, y)] '
Because f (X2, Y2) is a definite function value, the derivative of the second line is equal to 0
So the slope of the second line is [f (x, y) + F (X2, Y2)] = [f (x, y)] '

Given that the equation of line L is f (x, y) = 0, P1 (x1, Y1) and P2 (X2, Y2) are points on line L and outside line L respectively, then the equation f (x, y) - f (x1, Y1) - f (X2, Y2) = 0 denotes ()
A. A straight line passing through point P1 and perpendicular to l B. a straight line coincident with L C. a straight line passing through point P2 and parallel to l D. a straight line passing through point P2 but parallel to L

If the linear equation L is f (x, y) = 0, then the equation f (x, y) - f (x1, Y1) - f (X2, Y2) = 0, the two lines are parallel, P1 (x1, Y1) is the point on the linear L, f (x1, Y1) = 0, f (x, y) - f (x1, Y1) - f (X2, Y2) = 0, which is transformed into f (x, y) - f (X2, Y2) = 0. It is obvious that P2 (X2, Y2) satisfies the equation f (x, y) - f (x1, Y1) - f (X2, Y2) = 0 F (x, y) - f (x1, Y1) - f (X2, Y2) = 0 represents a straight line passing through P2 and parallel to L. therefore, C

The result is written as a power

=-(x-y)²(x-y)³(x-y)^(2n)
=-(x-y)^(2+3+2n)
=-(x-y)^(2n+5)

Expressed in the form of power of (X-Y): ((X-Y) to the 3rd power) to the 2nd power + ((Y-X) to the 3rd power) to the 2nd power + 2 (X-Y) to the 6th power=

Original formula = (X-Y) ^ 6 + (X-Y) ^ 6 + 2 (X-Y) ^ 6 = 4 (X-Y) ^ 6

The 2C power of E is equal to 4. What's the value of C

Logarithm of base 2 with e

The a power of 2 is equal to 3, the B power of 2 is equal to 6, and the C power of 2 is equal to 12,

2^a=3
2^b=6
2^c=12
2^(a+b-2c)=2^a*2^b/2^(2c)
=3*6/12^2
=18/144
=1/8
=2^(-3）
So, a + b-2c = - 3

It is known that the 2A power is equal to 3, the 2B power is equal to 6, and the 2C power is equal to 18

Solution
2^a=3
2^b=6
2^c=18
∴a=log2(3).b=log2(6).c=log2(18)
∵ y = log2 (x) is an increasing function
∴a