Question

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Answer 1

Solution :-

Taking LHS & simplifying

$\to\displaystyle\sf\dfrac{1}{\frac{1}{\frac{1}{\frac{1}{x}+\frac{1}{2}}+\frac{1}{\frac{1}{x}+\frac{1}{2}}}+ \frac{1}{\frac{1}{\frac{1}{x}+\frac{1}{2}}+\frac{1}{\frac{1}{x}+\frac{1}{2}}}}$

Write all the numerators above the least common denominator 2x .

$\to\displaystyle\sf\dfrac{1}{\frac{1}{\frac{1}{\frac{2+xr}{2x}}+\frac{1}{\frac{2+x}{2x}}}+ \frac{1}{\frac{1}{\frac{2+x}{2x}}+\frac{1}{\frac{2+x}{2x}}}}$

By simplifying the complex fraction

$\to\displaystyle\sf\dfrac{1}{\frac{1}{\frac{2x}{{2+x}}+\frac{2x}{{2+x}}}+ \frac{1}{\frac{2x}{{2+x}}+\frac{2x}{{2+x}}}}$

Adding like terms

$\to\displaystyle\sf\dfrac{1}{\frac{1}{2[\frac{2x}{{2+x}}]}+ \frac{1}{2[\frac{2x}{{2+x}}]}}$

$\to\displaystyle\sf\dfrac{1}{\frac{1}{\frac{4x}{{2+x}}}+ \frac{1}{\frac{4x}{{2+x}}}}$

$\to\displaystyle\sf\dfrac{1}{{\frac{2+x}{{4x}}}+ {\frac{2+x}{2x}}}$

Collecting the like terms,

$\to\displaystyle\sf\dfrac{1}{2\times\frac{2+x}{4x}}$

$\sf\to\dfrac{2x}{2+x}$

Now comparing with RHS

$\sf\to\dfrac{2x}{2+x}=\dfrac{x}{36}$

By cross-multiplying ,

$\sf\to x[2+x]=2x[36]$

$\sf\to 2x + x^2 = 72x$

$\sf\to x^2 - 70x = 0$

Taking common

$\sf\to x[x-70]=0$

$\sf\to x - 70 =c 0$

$\sf\to x = 70$

Hence, x = 70

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

Solve for x :

$\sf{\dfrac{1}{\dfrac{1}{\dfrac{1}{\tfrac{1}{x}+\tfrac{1}{2}}+\dfrac{1}{\tfrac{1}{x}+\tfrac{1}{2}}}+\dfrac{1}{\dfrac{1}{\frac{1}{x}+\frac{1}{2}}+\dfrac{1}{\frac{1}{x}+\frac{1}{2}}}}=\dfrac{x}{36}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{x=70}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\dfrac{1}{\dfrac{1}{\dfrac{1}{\tfrac{1}{x}+\tfrac{1}{2}}+\dfrac{1}{\tfrac{1}{x}+\tfrac{1}{2}}}+\dfrac{1}{\dfrac{1}{\frac{1}{x}+\frac{1}{2}}+\dfrac{1}{\frac{1}{x}+\frac{1}{2}}}}=\dfrac{x}{36}}$

$\sf{=\dfrac{1}{\dfrac{1}{\dfrac{1}{\frac{1}{x}+\frac{1}{2}}+\dfrac{1}{\frac{1}{x}+\frac{1}{2}}}+\dfrac{1}{\dfrac{4x}{x+2}}}}$

$\sf{=\dfrac{1}{\dfrac{1}{\dfrac{4x}{x+2}}+\dfrac{1}{\dfrac{4x}{x+2}}}}$

$\sf{=\dfrac{1}{\dfrac{x+2}{2x}}}$

$\sf{=\dfrac{2x}{x+2}}$

Cross multiply :

$\implies\sf{2x\cdot \:36=\left(x+2\right)x}$

$\implies\sf{72x=\left(x+2\right)x}$

$\implies\sf{72x-\left(x+2\right)x=\left(x+2\right)x-\left(x+2\right)x}$

$\implies\sf{72x-\left(x+2\right)x=0}$

$\implies\sf{-x\left(x-70\right)=0}$

Using the Zero Factor Principle:

If ab = 0 then a = 0 or b = 0 ( or both a = 0 and b = 0)

$\implies\sf{x=0}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$\sf{x=0,\:x=70}$

Since the equation is undefined for 0

$\large\boxed{\sf{x=70}}$