(10x+y)-(10y+x)=18 x=8-y

(10x+y)-(10y+x)=18 x=8-y

(10x+y)-(10y+x)=18 x=8-y
9x-9y=18
x-y=2
8-y-y=2
y=3
x=5

How to solve x + Y > 11,10x + y-18 = 10Y + X

∵x+y>11
10x+y-18=10y+x
=>x-y=2
That is, x = y + 2
∴y+2>11-y
y>9/2
x>13/2

In the quadratic function y = X2, if y decreases with the increase of X, what is the range of X

X is less than - 1 or X is greater than 1; or the absolute value of X is greater than 1

There is a three digit number, the number of its ten digit is equal to the sum of the number of its one digit and the number of its hundred digit, the number of its ten digit minus the number of its one digit is equal to 2, the number of its hundred digit and the number of its one digit are transposed, and the three digit number obtained is 99 larger than the original three digit number

If the three digit number is x, then the ten digit number is (x + 2) and the hundred digit number is 2, so 100x + 10 (x + 2) + 2 is 100x + 10 (x + 2) + 2 = 200 + 10 (x + 2) + 2 = 200 + 10 (x + 2) and the three digit number is x, then the ten digit number is (x + 2) and the hundred digit number is 2, so 100x + 10 (x + 2) is 2, so 100x + 10 (x + 10 (x + 2) + 2) is 2 = 100X (x (x + 10 (x + 2 + 2) + 2 = 200 + 10 (x + 10 (x + 10 (x + 2 + 2 + 2 + 3 (x + 2 + 2) is 2, so 100x + 100x + 10 (x + 10 (x + 10 (x + 3 (x + 3) is 2) is 2, the three three digit number of three digit number is x is x is x, so the three digit number of the three digit number of three digit number is x is x nbsp; X = 3 ten digit number is: x + 2 = 3 + 2 = 5, the three digit number is 253. A: the original three digit number is 253

6X to the second power - 11x-10 cross multiplication

The second power of 6x-11x-10
=（3x+2）（2x-5）
In the top right corner of my answer, click comment, and then you can choose satisfied, the problem has been solved perfectly

How to solve that 2.5x + 5 / 5 x equals 40.5

2.5x + 5 / 5 x equals 40.5
2.5x+0.2x=40.5
2.7x=40.5
x= 15

There is a point P (1,1) in the ellipse x23 + Y22 = 1. A straight line passes through the point P and intersects with the ellipse at P1 and P2. The chord p1p2 is bisected by the point P. the equation of the straight line p1p2 is obtained

Let P1 (x1, Y1) and P2 (X2, Y2), then x123 + y122 = 1 and x223 + y222 = 1 are subtracted to get (x1 + x2) (x1 − x2) 3 + (Y1 + Y2) (Y1 − Y2) 2 = 0 ∵ chord. P1p2 is bisected by point P, and then Y1 + x2 = 2 and Y1 + Y22 = 2 are substituted into the above formula to get Y1 − y2x1 − x2 = - 23, that is, the slope of line p1p2 is 23 ∵ the equation of line p1p2 is Y-1 = 23 (x-1), that is, 2x + 3y-5 = 0

The decimal number 25 is represented by 8421BCD code

0010 0101

Math class, the teacher out of a problem, calculation (1 / 2 + 1 / 3 + 1 / 4 +) +1/2013)×（1+1/2+1/3… 1/2012)-(1+1/2+1/3… 1/2013))×（1/2+1/3… 1/2012)

Let x = 1 / 2 + 1 / 3 + 1 / 4 +... 1 / 2012 (1 / 2 + 1 / 3 + 1 / 4 +... 1 / 2013) x (1 + 1 / 2 + 1 / 3 + 1 / 4 +... 1 / 2012) - (1 + 1 / 2 + 1 / 3 + 1 / 4 +... + 1 / 2013) x (1 / 2 + 1 / 3 + 1 / 4 +... + 1 / 2012) = (x + 1 / 2013) (1 + x) - (1 + X + 1 / 2013) x = x + 1 / 2013 + x-178; + X / 2013-x-178; - x

Let d be an interval in the domain of the function y = f (x). If there exists x0 ∈ d such that f (x0) = - x0, then x0 is said to be a subfixed of F (x)
If f (x) = ax2-3x-a+
5 / 2 has a fixed point in the interval [1,4]. Find the value range of constant a
The answer is (- ∞, 1 / 2]. Why can we get negative infinity?
Can let me understand, there are additional oh

If the function f (x) = ax2-3x-a + 5 / 2 has a fixed point in the interval [1,4], then ax & # 178; - 3x-a + 5 / 2 = - x, that is, ax & # 178; - 2x-a + 5 / 2 = 0 has a solution in [1,4], then a (X & # 178; - 1) = 2x-5 / 2. When x = 1, the equation does not hold. Let g (x) = (2x-5 / 2) / (X & # 178; - 1)