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


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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 deleteduser1 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 deleteduser1 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
	$x−(4/x)=2⇒ determinant ((((4/x)=x−2)))⇒ determinant ((((2/x)=((x−2)/2)))) ⇒ determinant (((x^2 =2x+4))) p=(x^2 /(x−2))−(x/(1+(2/x)))−(4/(1+(4/x^2 ))) =(x/(1−(2/x)))−(x/(1+(2/x)))−(4/(1+(4/x)∙(1/x))) =(x/(1−((x−2)/2)))−(x/(1+((x−2)/2)))−(4/(1+(x−2)((1/x)))) =(x/((2−x+2)/2))−(x/((2+x−2)/2))−(4/(1+((x−2)/x))) =(x/((4−x)/2))−(x/(x/2))−(4/(1+1−(2/x))) =((2x)/(4−x))−2−(4/(2−((x−2)/2))) =((2x)/(4−x))−2−(4/((6−x)/2)) =((2x)/(4−x))−(8/(6−x))−2 =((12x−2x^2 −32+8x)/(x^2 −10x+24))−2 =((−2x^2 +20x−32−2x^2 +20x−48)/(x^2 −10x+24)) =((−4x^2 +40x−96−80+96)/(x^2 −10x+24))=((−4(2x+4)+40x−80)/(2x+4−10x+24)) =((−8x−16+40x−80)/(−8x+28)) =((32x−96)/(−8x+28)) =((8x−24)/(−2x+7)) ▶(((32)/p))^2 =((32^2 )/p^2 )=((32^2 )/((64(x−3)^2 )/((−2x+7)^2 ))) =((32×32(−2x+7)^2 )/(64(x−3)^2 ))=((16(4x^2 −28x+49))/(x^2 −6x+9)) =((16{4(2x+4)−28x+49})/(2x+4−6x+9))=((16(−20x+65))/(−4x+13))=((16×5(−4x+13))/(−4x+13))=80$
	$x - \frac{4}{x} = 2 \Rightarrow \begin{matrix} \frac{4}{x} = x - 2 \end{matrix} \Rightarrow \begin{matrix} \frac{2}{x} = \frac{x - 2}{2} \end{matrix}$ $\Rightarrow \begin{matrix} x^{2} = 2 x + 4 \end{matrix}$ $p = \frac{x^{2}}{x - 2} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x^{2}}}$ $= \frac{x}{1 - \frac{2}{x}} - \frac{x}{1 + \frac{2}{x}} - \frac{4}{1 + \frac{4}{x} \cdot \frac{1}{x}}$ $= \frac{x}{1 - \frac{x - 2}{2}} - \frac{x}{1 + \frac{x - 2}{2}} - \frac{4}{1 + (x - 2) (\frac{1}{x})}$ $= \frac{x}{\frac{2 - x + 2}{2}} - \frac{x}{\frac{2 + x - 2}{2}} - \frac{4}{1 + \frac{x - 2}{x}}$ $= \frac{x}{\frac{4 - x}{2}} - \frac{x}{\frac{x}{2}} - \frac{4}{1 + 1 - \frac{2}{x}}$ $= \frac{2 x}{4 - x} - 2 - \frac{4}{2 - \frac{x - 2}{2}}$ $= \frac{2 x}{4 - x} - 2 - \frac{4}{\frac{6 - x}{2}}$ $= \frac{2 x}{4 - x} - \frac{8}{6 - x} - 2$ $= \frac{12 x - 2 x^{2} - 32 + 8 x}{x^{2} - 10 x + 24} - 2$ $= \frac{- 2 x^{2} + 20 x - 32 - 2 x^{2} + 20 x - 48}{x^{2} - 10 x + 24}$ $= \frac{- 4 x^{2} + 40 x - 96 - 80 + 96}{x^{2} - 10 x + 24} = \frac{- 4 (2 x + 4) + 40 x - 80}{2 x + 4 - 10 x + 24}$ $= \frac{- 8 x - 16 + 40 x - 80}{- 8 x + 28}$ $= \frac{32 x - 96}{- 8 x + 28}$ $= \frac{8 x - 24}{- 2 x + 7}$ $▸ {(\frac{32}{p})}^{2} = \frac{32^{2}}{p^{2}} = \frac{32^{2}}{\frac{64 {(x - 3)}^{2}}{{(- 2 x + 7)}^{2}}}$ $= \frac{32 \times 32 {(- 2 x + 7)}^{2}}{64 {(x - 3)}^{2}} = \frac{16 (4 x^{2} - 28 x + 49)}{x^{2} - 6 x + 9}$ $= \frac{16 {4 (2 x + 4) - 28 x + 49}}{2 x + 4 - 6 x + 9} = \frac{16 (- 20 x + 65)}{- 4 x + 13} = \frac{16 \times 5 (- 4 x + 13)}{- 4 x + 13} = 80$
	Commented by mathlove last updated on 10/Jan/24

	$s o g o o d$
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