Math, asked by manoswini, 1 year ago

$\frac{1}{x - a - b } = \frac{1}{x} - \frac{1}{a} - \frac{1}{b}$
Solve.....

Answers

Answered by abhi178

$\frac{1}{x-a-b}= \frac{1}{x} - \frac{1}{a} - \frac{1}{b} \\ \\ \frac{1}{x - a - b} - \frac{1}{x} = - \frac{1}{a} - \frac{1}{b} \\ \\ \frac{x - x + a + b}{ {x}^{2} - ax - bx } = \frac{ - (a + b)}{ab } \\ \\ \frac{1}{ {x}^{2} - ax - bx } = \frac{ - 1}{ab} \\ \\ {x}^{2} - (a + b)x + ab = 0 \\ \\ x = a \: \: and \: \: \: b$

manoswini: If my teacher gives this answer correct.... Then I shall thank you....

abhi178: its correct

abhi178: 100%

manoswini: Okay...... Let me see....

manoswini: Thanks a lot.....But you missed some of the steps which someone corrected after you.....

Anonymous: Thx!

Answered by Anonymous

Heya user,

$\frac{1}{x-a-b} = \: \frac{1}{x} \:-\: \frac{1}{a} \:-\: \frac{1}{b} \\ \frac{1}{x-a-b} - \frac{1}{x} = - \frac{1}{a} - \frac{1}{b} \\ \frac{x-(x-a-b)}{ x^{2} -ax-bx} = -\frac{a+b}{ab} \\ \frac{a+b}{ x^{2} -ax-bx} = -\frac{a+b}{ab} \\ x^{2} -ax-bx = -ab \\ x^{2} -ax-bx + ab =0 \\ x(x-a) - b(x-b) = 0 \\(x-a)(x-b)=0\\x=a,x=b;$

Hope your teacher remarks your answer... :)

manoswini: Thank you so much

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