((⋮BeMath⋮)/△) (((√x) +1)/(x(√x) +x+(√x))) : (1/( (√x) −x^2 )) + x = ?


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Question Number 108753 by bemath last updated on 19/Aug/20

$\frac{⋮ B e M a t h ⋮}{△}$ $\frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} : \frac{1}{\sqrt{x} - x^{2}} + x = ?$
Commented by Rasheed.Sindhi last updated on 19/Aug/20

$I^{'} v e t h o u g h t i t a s$ $\frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} : (\frac{1}{\sqrt{x} - x^{2}} + x)$ $b e c a u s e I t h i n k^{'} :^{'} h a s l o w p r i o r i t y$ $t h a n^{'} +^{'} (a c c o r d i n g t o c o n v e n t i o n)$ $- - - - - - - - - - - - - - -$ $\frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} : \frac{1}{\sqrt{x} - x^{2}} + x$ $\frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} : \frac{1 + x (\sqrt{x} - x^{2})}{\sqrt{x} - x^{2}}$ $\frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} \times \frac{\sqrt{x} - x^{2}}{1 + x \sqrt{x} - x^{3}}$ $\frac{(\sqrt{x} + 1) (\sqrt{x} - x^{2})}{(x \sqrt{x} + x + \sqrt{x}) (1 + x \sqrt{x} - x^{3})}$ $= \frac{\overset{\times}{(x - x^{2} + (x^{2} + 1) \sqrt{x})}}{\overset{\times}{(x - x^{2} + (x^{2} + 1) \sqrt{x})} (1 - x^{3} + x \sqrt{x})}$ $= \frac{1}{(1 - x^{3} + x \sqrt{x})} \times \frac{(1 - x^{3} - x \sqrt{x})}{(1 - x^{3} - x \sqrt{x})}$ $= \frac{(1 - x^{3} - x \sqrt{x})}{{(1 - x^{3})}^{2} - x^{3})} = \frac{(1 - x^{3} - x \sqrt{x})}{1 - 2 x^{3} + x^{6} - x^{3}}$ $= \frac{(1 - x^{3} - x \sqrt{x})}{x^{6} - 3 x^{3} + 1}$
Commented by bemath last updated on 19/Aug/20

Commented by bemath last updated on 19/Aug/20

$b u t n o t h i n g a n s w e r$
Commented by Rasheed.Sindhi last updated on 19/Aug/20

$Ok sir, a f t e r a l l {i t}^{'} s a m a t t e r o f$ $c o n v e n t i o n o n l y!$
	Answered by bemath last updated on 19/Aug/20

	Answered by 1549442205PVT last updated on 19/Aug/20

	$Put \frac{\sqrt{x} + 1}{x \sqrt{x} + x + \sqrt{x}} : \frac{1}{\sqrt{x} - x^{2}} + x = P$ $Since x \sqrt{x} + x + \sqrt{x} = \sqrt{x} (x + \sqrt{x} + 1)$ $\sqrt{x} - x^{2} = \sqrt{x} (1 - {(\sqrt{x})}^{3}) = \sqrt{x} (1 - \sqrt{x}) (1 + \sqrt{x} + x),$ $P = [\frac{\sqrt{x} + 1}{\sqrt{x} (1 + \sqrt{x} + x)} \times \sqrt{x} (1 - \sqrt{x}) (1 + \sqrt{x} + x)] + x$ $= (\sqrt{x} + 1) (1 - \sqrt{x}) + x = 1 - x + x = 1$ $Thus, P = 1$
	Commented by bemath last updated on 19/Aug/20

	$j o o s s$
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