if-p-x-2-x-2-x-1-2-x-4-1-4-x-2-and-x-4-x-2-then-show-that-32-p-2-80-

Question Number 203066 by mathlove last updated on 09/Jan/24

i f p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}} a n d x - \frac{4}{x} = 2

t h e n s h o w t h a t {(\frac{32}{p})}^{2} = 80

Answered by Rasheed.Sindhi last updated on 09/Jan/24

i f p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}} a n d x - \frac{4}{x} = 2

\underset{―}{t h e n s h o w t h a t {(\frac{32}{p})}^{2} = 80_{_{}}}

∙ p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}}

= \frac{x}{1 - \frac{2}{x}} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}}

= \frac{x (1 + \frac{2}{x}) (1 + \frac{4}{x^{2}}) - x (1 - \frac{2}{x}) (1 + \frac{4}{x^{2}}) - 4 (1 - \frac{2}{x}) (1 + \frac{2}{x})}{(1 - \frac{2}{x}) (1 + \frac{2}{x}) (1 + \frac{4}{x^{2}})}

= \frac{(x + 2) (1 + \frac{4}{x^{2}}) + (2 - x) (1 + \frac{4}{x^{2}}) - 4 (1 - \frac{2}{x}) (1 + \frac{2}{x})}{(1 - \frac{4}{x^{2}}) (1 + \frac{4}{x^{2}})}

= \frac{(1 + \frac{4}{x^{2}}) (x + 2 + 2 - x) - 4 (1 - \frac{2}{x}) (1 + \frac{2}{x})}{(1 - \frac{16}{x^{4}})}

= \frac{4 (1 + \frac{4}{x^{2}}) - 4 (1 - \frac{4}{x^{2}})}{\frac{x^{4} - 16}{x^{4}}}

= \frac{4 (1 + \frac{4}{x^{2}} - 1 + \frac{4}{x^{2}})}{\frac{x^{4} - 16}{x^{4}}}

= \frac{4 (\frac{8}{x^{2}})}{\frac{x^{4} - 16}{x^{4}}}

= \frac{32}{x^{2}} \times \frac{x^{4}}{x^{4} - 16}

= \frac{32 x^{2}}{x^{4} - 16}

= \frac{32}{x^{2} - \frac{16}{x^{2}}}

= \frac{32}{(x - \frac{4}{x}) (x + \frac{4}{x})}

= \frac{32}{(2) (x + \frac{4}{x})}

p = \frac{16}{x + \frac{4}{x}}

∙ p^{2} = \frac{16^{2}}{{(x + \frac{4}{x})}^{2}} = \frac{16^{2}}{{(x - \frac{4}{x})}^{2} + 16} = \frac{16^{2}}{{(2)}^{2} + 16} = \frac{16^{2}}{20}

▸ {(\frac{32}{p})}^{2} = \frac{32^{2}}{p^{2}} = \frac{32^{2}}{\frac{16^{2}}{20}} = 32^{2} \times \frac{20}{16^{2}} = 80

Answered by Rasheed.Sindhi last updated on 10/Jan/24

x - \frac{4}{x} = 2 \Rightarrow \begin{matrix} x^{2} = 2 x + 4 \end{matrix}

p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}}

= \frac{2 x + 4}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{2 x + 4}}

= \frac{2 (x + 2)}{x - 2} - \frac{x^{2}}{x + 2} - \frac{4}{\frac{2 x + 8}{2 x + 4}}

= \frac{2 (x + 2)}{x - 2} - \frac{2 x + 4}{x + 2} - \frac{4 (x + 2)}{x + 4}

= \frac{2 (x + 2)}{x - 2} - \frac{4 (x + 2)}{x + 4} - 2

= \frac{2 (x + 2) (x + 4) - 4 (x + 2) (x - 2)}{(x - 2) (x + 4)} - 2

= \frac{2 x^{2} + 12 x + 16 - 4 x^{2} + 16}{x^{2} + 2 x - 8} - 2

= \frac{- 2 x^{2} + 12 x + 32}{x^{2} + 2 x - 8} - 2

= \frac{- 2 x^{2} + 12 x + 32 - 2 x^{2} - 4 x + 16}{x^{2} + 2 x - 8}

= \frac{- 4 x^{2} + 8 x + 48}{x^{2} + 2 x - 8}

= \frac{- 4 (2 x + 4) + 8 x + 48}{2 x + 4 + 2 x - 8}

= \frac{- 8 x - 16 + 8 x + 48}{4 x - 4} = \frac{32}{4 (x - 1)} = \frac{8}{x - 1}

▸ {(\frac{32}{p})}^{2} = {(\frac{32}{\frac{8}{x - 1}})}^{2} = {(\frac{32 (x - 1)}{8})}^{2}

= 16 (x^{2} - 2 x + 1)

= 16 (2 x + 4 - 2 x + 1) = 16 (5) = 80

Commented by mathlove last updated on 09/Jan/24

o k t h a n k s

Answered by AST last updated on 09/Jan/24

x^{2} - 4 = 2 x \Rightarrow (x - 2) (x + 2) = 2 x

{(x - 1)}^{2} - 5 = 0 \Rightarrow x = 1 \underline{+} \sqrt{5}

\Rightarrow p = \frac{x}{\frac{2}{x + 2}} - \frac{x}{\frac{2}{x - 2}} - \frac{4}{\frac{(x + 4)}{(x + 2)}} = \frac{x (x + 2) - x (x - 2)}{2} - \frac{4 (x + 2)}{x + 4}

= 2 x - \frac{4 (x + 2)}{x + 4} = \frac{2 (x^{2} + 2 x - 4)}{x + 4} = \frac{8 x}{x + 4}

{(\frac{32}{p})}^{2} = {[\frac{32 (x + 4)}{8 x}]}^{2} = {[\frac{4 (x + 4)}{x}]}^{2} = {(4 + \frac{16}{x})}^{2} = {(\underline{+} 4 \sqrt{5})}^{2} = 80

Answered by AST last updated on 09/Jan/24

p = \frac{x^{2}}{x - 2} - \frac{x^{2}}{x + 2} - \frac{4 x^{2}}{x^{2} + 4} = 4 x^{2} (\frac{1}{x^{2} - 4} - \frac{1}{x^{2} + 4})

= \frac{32 x^{2}}{(x^{2} - 4) (x^{2} + 4)} = \frac{32 x^{2}}{2 x (2 x + 8)} = \frac{8 x}{x + 4}

\Rightarrow {(\frac{32}{p})}^{2} = {[\frac{4 (x + 4)}{x}]}^{2} = {(\underline{+} 4 \sqrt{5})}^{2} = 80

Answered by Rasheed.Sindhi last updated on 09/Jan/24

x - \frac{4}{x} = 2 \Rightarrow \begin{matrix} \frac{4}{x} = x - 2 \end{matrix} \Rightarrow \begin{matrix} \frac{2}{x} = \frac{x - 2}{2} \end{matrix}

\Rightarrow \begin{matrix} x^{2} = 2 x + 4 \end{matrix}

p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}}

= \frac{x}{1 - \frac{2}{x}} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x} \cdot \frac{1}{x}}

= \frac{x}{1 - \frac{x - 2}{2}} - \frac{x}{1 + \frac{x - 2}{2}} - \frac{4}{1 + (x - 2) (\frac{1}{x})}

= \frac{x}{\frac{2 - x + 2}{2}} - \frac{x}{\frac{2 + x - 2}{2}} - \frac{4}{1 + \frac{x - 2}{x}}

= \frac{x}{\frac{4 - x}{2}} - \frac{x}{\frac{x}{2}} - \frac{4}{1 + 1 - \frac{2}{x}}

= \frac{2 x}{4 - x} - 2 - \frac{4}{2 - \frac{x - 2}{2}}

= \frac{2 x}{4 - x} - 2 - \frac{4}{\frac{6 - x}{2}}

= \frac{2 x}{4 - x} - \frac{8}{6 - x} - 2

= \frac{12 x - 2 x^{2} - 32 + 8 x}{x^{2} - 10 x + 24} - 2

= \frac{- 2 x^{2} + 20 x - 32 - 2 x^{2} + 20 x - 48}{x^{2} - 10 x + 24}

= \frac{- 4 x^{2} + 40 x - 96 - 80 + 96}{x^{2} - 10 x + 24} = \frac{- 4 (2 x + 4) + 40 x - 80}{2 x + 4 - 10 x + 24}

= \frac{- 8 x - 16 + 40 x - 80}{- 8 x + 28}

= \frac{32 x - 96}{- 8 x + 28}

= \frac{8 x - 24}{- 2 x + 7}

▸ {(\frac{32}{p})}^{2} = \frac{32^{2}}{p^{2}} = \frac{32^{2}}{\frac{64 {(x - 3)}^{2}}{{(- 2 x + 7)}^{2}}}

= \frac{32 \times 32 {(- 2 x + 7)}^{2}}{64 {(x - 3)}^{2}} = \frac{16 (4 x^{2} - 28 x + 49)}{x^{2} - 6 x + 9}

= \frac{16 {4 (2 x + 4) - 28 x + 49}}{2 x + 4 - 6 x + 9} = \frac{16 (- 20 x + 65)}{- 4 x + 13} = \frac{16 \times 5 (- 4 x + 13)}{- 4 x + 13} = 80

Commented by mathlove last updated on 10/Jan/24

s o g o o d