If 1p - 1q - 1x - 1x - p - q and 1p - 1q - 1x - 1x - 1x - p - q - 0- then find the value of x-

Given, 1p+1q+1x=1x+p+q ....(1) (1p+1q #160; )+1x (1x 1x+p+q ) ....(2)(2) becomes 1p+1q+1x 1x+ 1x+p+q=0 ⇒ ( 1p+1q )=1x+p+q ...(3)equating ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Law of Reciprocal

If 1p + 1q ...

Question

If $\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x + p + q}$ and $(\frac{1}{p} + \frac{1}{q}) + \frac{1}{x} - (\frac{1}{x} - \frac{1}{x + p + q}) = 0$ , then find the value of $x .$

Open in App

Solution

Given,

$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x + p + q}$ ....(1)

$(\frac{1}{p} + \frac{1}{q}) + \frac{1}{x} - (\frac{1}{x} - \frac{1}{x + p + q})$ ....(2)

(2) becomes

$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} - \frac{1}{x} + \frac{1}{x + p + q} = 0$

$\Rightarrow - (\frac{1}{p} + \frac{1}{q}) = \frac{1}{x + p + q}$ ...(3)

equating (1) and (3) as both have same RHS

$\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = - (\frac{1}{p} + \frac{1}{q})$

$\frac{2}{p} + \frac{2}{q} + \frac{1}{x} = 0$

$\frac{1}{x} = - 2 (\frac{1}{p} + \frac{1}{q}) = - 2 (\frac{p + q}{p q})$

$\Rightarrow x = - \frac{1}{2} (\frac{p q}{p + q})$

Suggest Corrections

Similar questions

Q. If r,s are the roots of the equation x²-px+q=0, then what is the equation of the curve whose roots are ([1/r] + s) and ([1/s]+r)?

If $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x =_________.$

Q. If

{sin}^{- 1} {\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}} = \frac{π}{p} + \frac{{cos}^{- 1} x}{q}

0 < x < 1

Find the value of

p + q

Q. If

y = \frac{1}{1 + x^{q - p} + x^{r - p}} + \frac{1}{1 + x^{r - q} + x^{p - q}} + \frac{1}{1 + x^{p - r} + x^{q - r}}

then

\frac{d y}{d x} =

Q. If

P = x^{3} - \frac{1}{x^{3}}

and

Q = x - \frac{1}{x}, x \in (0, x)

then minimum value value of

\frac{P}{Q}

Join BYJU'S Learning Program

If 1p+1q+1x=1x+p+q and (1p+1q)+1x−(1x−1x+p+q)=0, then find the value of x.

Given,1p+1q+1x=1x+p+q....(1)(1p+1q)+1x−(1x−1x+p+q)....(2)(2) becomes1p+1q+1x−1x+1x+p+q=0⇒−(1p+1q)=1x+p+q...(3)equating (1) and (3) as both have same RHS1p+1q+1x=−(1p+1q)2p+2q+1x=01x=−2(1p+1q)=−2(p+qpq)⇒x=−12(pqp+q)

If $\frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{x + p + q}$ and $(\frac{1}{p} + \frac{1}{q}) + \frac{1}{x} - (\frac{1}{x} - \frac{1}{x + p + q}) = 0$ , then find the value of $x .$