It is known that the difference between two natural numbers is & nbsp; 48, and their least common multiple is & nbsp; 60______ And______ ．

It is known that the difference between two natural numbers is & nbsp; 48, and their least common multiple is & nbsp; 60______ And______ ．

Let two natural numbers be a, B, and a ＞ B1. A and B are coprime, then AB = 60, and A-B = 48, so a (a-48) = 60. The solution shows that a and B are irrational numbers, which contradicts the condition. Therefore, a and B cannot be coprime. 2. A and B are not coprime. (1) a is a multiple of B, then a = 60, B = 60-48 = 12 (2) a is not a multiple of B. let Ma = NB = 60 = 2 × 2 × 3 × 5, that is, Ma = n (a-48) = 2 × 2 × 3 × 5, and a and (a-48) are divisors of 60, Then a may be 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, a is more than 48, a can only be (1) cases, so a = 60, B = 12

How to solve the equation 5x - one fifth x = 19.2

5x - one fifth x = 19.2
4.8X=19.2
X=19.2/4.8
X=4

Is there a constant k such that the two real roots ab of equation 9 times the square of x minus (4k-7) - 6 times the square of k = 0 satisfy | A / b | = 3 / 2? If so, K is requested; if not, please explain the reason

The equation for X is like this
9x^2 -(4k-7)x+6k^2=0
If it exists, it must satisfy Δ > 0, that is, (4k-7) ^ 2 + 4 × 9 × 6K ^ 2 > 0
Find out the range of K
Because, two roots a / b = 3 / 2, obviously two are not 0, and the same sign AB > 0,
According to Weida's theorem, a + B = (4k-7) / 9, ab = 2K ^ 2 / 3
Then (a ^ 2 + B ^ 2) / AB = [(a + b) ^ 2-2ab] / AB = A / B + B / A
={[(4k-7)/9]^2 – 2×(2k^2 /3)}/ (2k^2 /3)=3/2+2/3
Through this formula to find out the value of K, see if it is in the above range!
If it is, it exists; if it is not, it does not exist

Given that the square of (x + y) is 3 and the square of (X-Y) is 7, then the value of [(XY + 2) (XY-2) - 2x's Square, the square of Y + 4] / one-half of XY is?

(X+Y)^2=X^2+2XY+Y^2=3
(X-Y)^2=X^2-2XY+Y^2=7
subtract
4XY=-1
XY=-1
[(XY+2)(XY-2)-2X^2Y^2+4]÷(XY/2)
=(X^2Y^2-4-2X^2Y^2+4)*2/(XY)
=-X^2Y^2*2/(XY)
=-2XY
=-2*(-1)
=2

To solve the equations: 1 / 4 square + XY + y square = 1 / 4 (x + y) square - 4 (x + y) - 12 = 0 (2)

From (1) we get (x / 2 + y) &# 178; = (1 / 2) &# 178; ｜ X / 2 + y = 1 / 2 x / 2 + y = - 1 / 2 from (2) we get (x + y-6) (x + y + 2) = 0 ｜ x + y-6 = 0 x + y + 2 = 0 ｜ X / 2 + y = 1 / 2 x / 2 + y = - 1 / 2 x / 2 + y = - 1 / 2x + y-6 = 0 x + y-6 = 0 x + y + 2 = 0 ｜ x = 11 x = - 5 x = 13 x = - 5 y = 3 y

When the value of integer a is, the solution XY of the system of equations x + y = - 2,5x + 3Y = 2A is negative

x. Y should be greater than - 2 and less than 0
5x+3y=2a
-6+2x=2a
x=a+3
A + 3 greater than - 2 less than 0
So a is greater than - 5 and less than - 3
So a equals - 4

Finding the integer solution of the system of equations X-Y = XY

xy=x-y
xy-x+y-1=-1
x(y-1)+(y-1)=-1
(x+1)(y-1)=-1
Integer solution required
so
X + 1 = 1, Y-1 = - 1 or x + 1 = - 1, Y-1 = 1
Solution
X = 0, y = 0 or x = - 2, y = 2

Finding the integer solution of the system of equations 2|x| + y = 7, y = |x| + 2

y=|x|+2
2|x|+y=7
Then 2 | x | + | x | + 2 = 7
3|x|=5
|x|=5/3
X is not an integer
So there is no integer solution
The solution is
x=5/3,y=11/3
x=-5/3,y=11/3

If the absolute value of 2 (x-1) + X + 2Y + 1 = 0, then what is XY of 1 / 2

Because 2 (x-1) ^ 2 is greater than or equal to 0,
The absolute value of X + 2Y + 1 is greater than or equal to 0
SO 2 (x-1) ^ 2 = 0
x+2y+2=0
So x = 1
1+2y+2=0
y=-3/2
So - XY / 2 = 1 * (- 3 / 2) / - 2 = 3

When k is a value, the power of the absolute value of 2k-1 of the polynomial 4x multiplied by Y + XY-5 is a quartic polynomial? What is the power of X?

4x ^| 2k-1 × (y + XY-5) = 4x ^| 2k-1 | y + 4x ^ (| 2k-1 | + 1) Y-5 if the polynomial is a quartic polynomial, then | 2k-1 | + 1 + 1 = 4, k = 3 / 2 or K = - 1 / 2, so | 2k-1 | + 1 = 3 is a cubic expression of X