Find the natural numbers a and B satisfying the following three conditions at the same time 1. A is greater than B, 2, AB / A + B = 169, 3, a + B are complete square numbers

A = 28730,B = 170
Let a + B = m ^ 2
Then a = m ^ 2-B
A×B
= 169×(A+B)
= 169×M^2
Again
A×B
=(M^2-B)×B
=M^2×B - B^2
Simultaneous access:
M^2×B - B^2 = 169×M^2
Write a quadratic equation of one variable about B
B^2 - M^2×B +169×M^2= 0
According to the condition of integer root
(- M^2)^2 - 4×(169×M^2)
=M^4 - 676×M^2
=M^2 × (M^2 -676) ≥ 0
Square root of M ≥ 676
That is, m ≥ 26 and m ^ 2 - 676 is a complete square number
The solution is m = 170. (when m = 26, B has only one solution, and a = B discards it.)
The solution of B for M = 170
=[- (- m ^ 2) ± m ^ 2 × (m ^ 2 - 676)] / 2
B1 = 28730, corresponding to A1 = 170, A1

Find the satisfied conditions: √ (A-2 √ b) = √ X - √ y in the value of natural numbers a, x, y

Square on both sides
a-2√b=x-2√xy+y
So x + y = a
xy=b

Given that a / 8 + B / 16 = 1 / 2 (A and B are non-zero natural numbers), what are a and B respectively? How many groups of different answers can you write that meet the above conditions?
A / 8 is a eighth, B / 16 is B sixteenth

(2a+b)=8
b=2 a=3
b=4 a=2
b=6 a=1
There are three groups above

Given that the equation 1 / X-2 + K / x + 2 = 3 / x ^ 2-4 about X has no solution, find the value of K

If you go to the denominator, you get:
(x＋2)＋k(x－2)=3
(k＋1)x=2k＋1
Then when k = - 1, there is no solution; or when the root of the equation is 2 or - 2 [that is, the increasing root of the original equation], there is no solution, and the solution is k = - 3 / 4
Then when k = - 1 or K = - 3 / 4, the original equation has no solution

The known equation (k2-1) x2 + (K + 1) x + (k-7) y = K + 2= When k, the equation is linear= &The equation is a quadratic equation of two variables

Since the equation does not indicate whether it is about X or Y, we should refer to the equation of first degree with respect to X or y. when k2-1 = 0 and K + 1 = 0, the equation is the equation of first degree with respect to y, and the solution is k = - 1. When k2-1 = 0 and K + 1 ≠ 0, the equation is the equation of second degree with respect to y, and the solution is k = 1. So the answer is: - 1,1

When a is a value, the fractional equation 4 / (x-3) + 5 / (x + 3) = ax / (x ^ 2-9) has an increasing root

Multiply by (x-3) (x + 3)
4(x+3)+5(x-3)=ax
9x-3=ax
x(9-a)=3
x=3/(9-a)
The reason for increasing roots is that the denominator is equal to 0, that is, x = 3 or - 3
3/(9-a)=3 9-a=1 a=8
3/(9-a)=-3 9-a=-1 a=10

It is known that the fractional equation of x 3 + X + 1 / ax + 3 = 2 has an increasing root. How to find the value of A. (x-1 / X & # 178; - 1) and subtract 1

If there is an increasing root, then the increasing root is x = 0 or x = - 1
Remove the denominator of the original equation to get 3 (x + 1) + X (AX + 3) = 2x (x + 1)
It does not hold when x = 0
When x = - 1, - 1 (- A + 3) = 0, a = 3
(X & # 178; - 1) - 1 = x + 1-1 = x

The fractional equation AX = 2-x + 3 / 1 has an increasing root x = - 1. Find the value of A

Double x (x + 1)
3(x+1)+ax²=2x(x+1)+3x
If x = - 1 is an increasing root, then x = - 1 is the root of the equation
So 0 + a = 0-3
a=-3

If the fractional equation (AX / X-2) = (4 / X-2) - 2 has an increasing root 2, then the value of a is

A = 2, right

Let m be an integer, and M is not equal to 0, and the equation MX ^ 2 - (m-1) x + 1 = 0 has only rational root, so we can find the value of M
Such as the title

The equation MX ^ 2 - (m-1) x + 1 = 0 has only rational roots, and the discriminant = (m-1) ^ 2-4m = m ^ 2-6m + 1. Because the equation has only rational roots, the discriminant must be a complete square. Let m ^ 2-6m + 1 = k ^ 2, that is: m ^ 2-6m + 9-k ^ 2 = 8 (M-3) ^ 2-k ^ 2 = 8 (M-3 + k) (m-3-k) = 8 = 1 * 8 = 2 * 4, so: M-3 + k = 1 or 8, m-3-k = 8

Find the natural numbers a and B satisfying the following three conditions at the same time 1. A is greater than B, 2, AB / A + B = 169, 3, a + B are complete square numbers