Question

Divide 50 into two parts, such that the sum of their reciprocals is

1/1​

Answer 1

Explanation:

Let X and Y are two numbers.

according to question x+y =50

• Y= 50-x

● Now,

$= > \frac{1}{x} + \frac{1}{y} = \frac{1}{1} \\ \\ = > \frac{1}{x} + \frac{1}{50 - x} = \frac{1}{1} \\ \\ = > \frac{1(50 - x) + 1(x)}{x(50 - x)} = \frac{1}{1} \\ \\ = > \frac{50 - x + x}{50x - x ^{2} } = \frac{1}{1} \\ \\ = > \frac{50}{50x - {x}^{2} } = \frac{1}{1} \\ \\ = > 50x - {x}^{2} = 50 \\ \\ = > - {x}^{2} + 50x - 50 = 0 \\ \\ = > {x}^{2} - 50x + 50 = 0$

By solving this you will get..

x=25+5√23 or x=25−5√23

I hope it's helps you...

Please mark it as brainliest....

Answer 2

Explanation:

Let X and Y are two numbers.

according to question x+y =50

• Y= 50-x

● Now,

$= > \frac{1}{x} + \frac{1}{y} = \frac{1}{1} \\ \\ = > \frac{1}{x} + \frac{1}{50 - x} = \frac{1}{1} \\ \\ = > \frac{1(50 - x) + 1(x)}{x(50 - x)} = \frac{1}{1} \\ \\ = > \frac{50 - x + x}{50x - x ^{2} } = \frac{1}{1} \\ \\ = > \frac{50}{50x - {x}^{2} } = \frac{1}{1} \\ \\ = > 50x - {x}^{2} = 50 \\ \\ = > - {x}^{2} + 50x - 50 = 0 \\ \\ = > {x}^{2} - 50x + 50 = 0$

By solving this you will get..

x=25+5√23 or x=25−5√23

I hope it's helps you...

Please mark it as brainliest....