Two natural numbers, a and B, satisfy two conditions at the same time: 1 / 7 is less than 1 / 6 of B, a is less than 1 / 6 and a + B = 22

a=3 b=19

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AB is two non-zero natural numbers, they satisfy the following two conditions: 1, 7 / 1 is greater than a / B is greater than 6 / 1, 2, a + B = 22. Find the value of a and B,

1 / 7 < A / b < 1 / 6?
from
1/7

Both a and B are non-zero natural numbers. They satisfy the following two conditions: ① 1 / 7 ＜ A / b ＜ 1 / 6; ② a + B = 22?

n/7n＜A/B＜n/6n
6n+k+n=22,(k＜n)
n=3,k=1
A=3,B=19

It is known that the equation X-7 = K + 1 and 4k-x = 2-2x about X have the same solution

x=k+8,x=2-4k
So K + 8 = 2-4k
k=-(5/6)

If the equation 2 / x + 1 + 3 / X-1 = K / X * 2-1 has no solution, what is the value of K?

We only need to let x = 1 or - 1, then the value of K in the equation is meaningless. So when the value of K is 6 or - 4, the equation has no solution

When k is a value, the equation (X-2) / (x + 2) + (k) / (x ^ 2-4) = (x + 2) / (X-2) has no solution

Equation (X-2) / (x + 2) + K / (X & # 178; - 4) = (x + 2) / (X-2)
Multiply both sides by X & # 178; - 4, i.e. (x + 2) (X-2)
（x-2)²+k=(x+2)²
Simplify
8x=k
x=k/8
If the original equation has no solution, then x = K / 8 is the increasing root of the equation
Then K / 8 = 2 or K / 8 = - 2
‖ k = 16 or K = - 16
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If the equation KX − 3 + 2 = 4 − XX − 3 about X has no solution, then the value of K is ()
A. 3B. 1C. -1D. 0

The denominator of the equation is removed to get K + 2 (x-3) = 4-x. substituting x = 3 from the meaning of the question to get k = 4-3 = 1

If the equation 1 / (X-2) + K / (x + 2) = 3 / (x + 2) (X-2) about X has no solution, find the value of K

1/(x-2)+k/(x+2)=3/(x+2)(x-2)
Multiply both sides (x + 2) (X-2)
x+2+k(x-2)=3
x+kx-2k=1
There is no solution to cause
so x-2=0
or x+2=0
Get: x = 2 or x = - 2
When k is any number, the equation has no solution

If the equation X-1 / 6 + X / 3 = x ^ 2-x / x + K has no solution, find the value of fraction 9-k ^ 2 / K ^ 2-1-k + 3 / 2K + 1

Predigested fraction
（k2-1）/(9-k2)+（2k+1）/(k+3)
= (1-k2）/(k2-9)+ [（2k+1）(k-3)]/(k2-9)
= ( k2-5k-2)/(k2-9)
The equation is (9x-3) / (x ^ 2-x) = (x + k) / (x ^ 2-x)
The equation has no solution and the increasing root is x = 0 or x = 1
Let 9x-3 = x + K
When x = 0, k = - 3, the original fraction is meaningless
When x = 1, k = 5, the original fraction = - 1 / 8

If there is no solution to the equation 1 / (X-2) + K / (LX + 2) = 3 / X square-4 of X, find the value of K

1 / X-2 + K / x + 2 = (x + 2 + kx-2k) / (x-4) = 3 / (x-4) x + 2 + kx-2k = 3. When x = ± 2, the equation has no solution. When x = 2, x + 2 + kx-2k = 4 ≠ 3. When x = - 2, x + 2 + kx-2k = - 4k-4k = 3, k = - 3 / 4. So when k = - 3 / 4, the equation has no solution

Two natural numbers, a and B, satisfy two conditions at the same time: 1 / 7 is less than 1 / 6 of B, a is less than 1 / 6 and a + B = 22