Question

the term independent of x in the expansion of
$\sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x}$
Solve it.
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Answer 1

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HERE IS UR ANSWER

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$\sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x}$

$\frac{2x \sqrt{x} - 3 }{2x \sqrt{3} }$

$\frac{2 {x}^{ \frac{3}{2}} - 3}{2x \sqrt{3} }$

Answer 2

Step-by-step explanation:

$\bf\sqrt{ \frac{x}{3} } - \frac{ \sqrt{3} }{2x} \\ \\ \bf = > \frac{ \sqrt{x} }{ \sqrt{3} } - \frac{ \sqrt{3} }{2x}$

$\bf = > \dfrac{( \sqrt{x} \times 2x) - ( \sqrt{3} \times \sqrt{3} ) }{ \sqrt{3} \times 2x} \tiny{..taking \: lcm \: as \: \sqrt{3} \times 2x}$

$\\ \large \bf = > \frac{2x \sqrt{x} - 3}{2x \sqrt{3} }$

If you want to write in exponential form, you can write -

$\large \bf \frac{2 {x}^{ \frac{3}{2} } - 3}{2x {3}^{ \frac{1}{2} } }$

Hope it helps.

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