If 3 / 17 of the reciprocal of a is equal to 2 / 9 of the reciprocal of B, then which is the largest of a and B? (A and B are non-zero natural numbers

If 3 / 17 of the reciprocal of a is equal to 2 / 9 of the reciprocal of B, then which is the largest of a and B? (A and B are non-zero natural numbers

A's score is 3-17, B's score is 2-9, a's score is 5-2 / 3, B's score is 4-1 / 2, so a is bigger than B

If the natural numbers a and B satisfy the following conditions: 1 / A-1 / b = 1 / 195, a: B = 3:5, then a + B = ()

From a: B = 3:5, let a be equal to 3x, then B is equal to 5x
1/A-1/B=1/195
1/3x-1/5x=1/195
5 / 15x-3 / 15x = 1 / 195
8 / 15x = 1 / 195, x = 104
So a = 3 * 104 = 312, b * 104 = 520, so a + B = 312 + 520 = 832

If a is 18, then B + C = ()

Fill in: 8
A = 18, then
b=18x1/3=6
c=6x1/3=2
b+c=8

If the natural number AB satisfies 1 / A-1 / b = 1 / 195 and a; b = 3:5, then a + B=

∵A;B=3:5
Let a = 3x, B = 5x
∴1/(3x) -1/(5x)=1/195
5/(15x)-3/(15x)=1/195
2/(15x)=1/195
390=15x
∴x=26
∴A+B=3x+5x=8x=8×26=208

Known: a + B = 5 - √ 3, a is a natural number, 0 < B < 1, find 2A square - b square - b square one

Because 1

Given that a and B are all integers less than 100, and that 3 / A and 2 / a plus 1 are continuous natural numbers, what is the number pair (a, b)?

Let B / 5 = n, then a / 3 = n-1, (a + 1) / 2 = n + 1
a=3n-3
a+1=2n+2,
3n-3+1=2n+2,n=4.
a=9,b=20
There is only one pair of numbers (a, b) (9, 20)

a. B two natural numbers, they satisfy the following two conditions at the same time: (1) 1 / 7

（1）1/7

a. B, two natural numbers, they satisfy two conditions: 1 / 7 is less than a / B is less than 1 / 6, a + B = 22, find the value of a, B

From the inequality, we can calculate the range of a and B. one of them is greater than 18 and less than 20, but it is a natural number, and a + B = 22. Therefore, one of them is 19 and the other is 3

Given that the equation 4K (x + 2) - 1 = 2x about X has no solution, find K

4k(x+2)-1=2x
4kx+8k-1=2x
4kx-2x=1-8k
(4k-2)x=1-8k
∵ the equation 4K (x + 2) - 1 = 2x on X has no solution
∴4k-2=0,1-8k≠0
∴k=½

The equation x ^ 2-2x (K-3) x + K ^ 2-4k-1 = 0 for X is known
1) If the equation has a real root, find the range of K. 2) if the equation has a root 1, find the value of K. 3) if the two roots of the equation x ^ 2-2x (K-3) x + K ^ 2-4k-1 = 0 are taken as abscissa and ordinate points, which are just on the image of inverse scale function y = m / x, find the minimum value of M satisfying the condition
The third problem is: if the two roots of the equation x ^ 2-2 (K-3) x + K ^ 2-4k-1 = 0 are taken as abscissa and ordinate points just on the image of inverse scale function y = m / x, the minimum value of M satisfying the condition can be obtained.

1）△=4(k^2-6k+9)-4k^2+16k+4=40-8k≥0
Then the range of K is k ≤ 5
2) Substituting x = 1 into the original formula, the solution is k = 3 + radical 3 or K = 3-radical 3
3) That is, x1x2 = m, that is, K ^ 2-4k-1 = m, M = (K-2) ^ 2-5, that is, the minimum value of M is - 5