Question

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Answer 1

Answer:

REFER THE ABOVE ATTACHMENT

Step-by-step explanation:

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Answer 2

⠀⌬⠀ Find the Value of  ❝ x + y + xy ❞  if ,

$\qquad \bigstar \:\:\sf \bf{x} \:\sf {=\: \dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:}\:\&\:\bf{y}\:=\:\sf{\dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}}\:\:\:.\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dashrightarrow \:\:\sf \bf{x} \:\sf{ =\: \dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:\:\&\:}\:\bf{y}\:\sf{=\:\dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:\:}\:\:$

$\qquad \bigstar \:\:\pmb{\sf{\purple {\underline{ By \:Rationalizing \:the \:Given \:Value \:of \:x \:and \:y \::\:}}}}$

$\qquad \dashrightarrow \sf \:\: x \:=\: \dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:\:$

$\qquad \dashrightarrow \sf \:\: x \:=\: \dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:\times \:\dfrac{\sqrt{5}\:+\:\sqrt{3}}{\sqrt{5}\:+\:\sqrt{3}}\:\:$

$\qquad \dashrightarrow \sf \:\: x \:=\: \dfrac{\big( \sqrt{5}\:\big)^2 \:+\:2\sqrt{5}\times \sqrt{3} +\:\big( \sqrt{3}\big)^2}{\big( \sqrt{5}\big)^2\:-\:\big( \sqrt{3}\big)^2}\:\:$

$\qquad \dashrightarrow \sf \:\: x \:=\: \dfrac{ 5 \:+\:2\sqrt{15} +\:3 }{ 5\:-\:3 }\:\:$

$\qquad \dashrightarrow \sf \:\: x \:=\: \dfrac{ 8 \:+\:2\sqrt{15} }{ 2 }\:\:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple {\:\: x \:=\: 4 \:+\:\sqrt{15} \:}}}}}\:\:\bigstar \:$

$\qquad \dashrightarrow \sf \:\: y \:=\: \dfrac{\sqrt{5}\:-\:\sqrt{3}}{\sqrt{5}\:+\:\sqrt{3}}\:\:$

$\qquad \dashrightarrow \sf \:\: y \:=\: \dfrac{\sqrt{5}\:-\:\sqrt{3}}{\sqrt{5}\:+\:\sqrt{3}}\:\times \:\dfrac{\sqrt{5}\:-\:\sqrt{3}}{\sqrt{5}\:-\:\sqrt{3}}\:\:$

$\qquad \dashrightarrow \sf \:\: y \:=\: \dfrac{\big( \sqrt{5}\:\big)^2 \:-\:2\sqrt{5}\times \sqrt{3} +\:\big( \sqrt{3}\big)^2}{\big( \sqrt{5}\big)^2\:-\:\big( \sqrt{3}\big)^2}\:\:$

$\qquad \dashrightarrow \sf \:\: y \:=\: \dfrac{ 5 \:-\:2\sqrt{15} +\:3 }{ 5\:-\:3 }\:\:$

$\qquad \dashrightarrow \sf \:\: y \:=\: \dfrac{ 8 \:-\:2\sqrt{15} }{ 2 }\:\:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple {\:\: y \:=\: 4 \:-\:\sqrt{15} \:}}}}}\:\:\bigstar \:$

Therefore ,

$\qquad \bigstar \:\:\pmb{\sf{x\:=\:}\pink{ \: 4 \:+\:\sqrt{15}\:}\:\:\:\&\:\:\:\:\sf{y\:=\:}\pink {\:4 \:-\:\sqrt{15}\:\:\:.}}\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀¤ Finding Value of ❝ x + y + xy ❞ :

$\qquad:\implies \sf x + y + x y \:$

$\qquad \dag\:\underline {\frak{ Substituting \:known \:Values \:\::\:}}$

$\qquad:\implies \sf \Big\{ \big( 4 \:+\:\sqrt{15} \big) + \big( 4 \:-\:\sqrt{15} \big) \Big\} + \Big\{ \big(4 \:+\:\sqrt{15}\big)\:\big( 4 \:-\:\sqrt{15}\big)\:\Big\} \:\\\\\\ \qquad:\implies \sf \Big\{ 4 \:+\:\sqrt{15} + 4 \:-\:\sqrt{15} \Big\} + \Big\{ \big( 4\big)^2\:-\:\big( \sqrt{15}\big)^2\:\Big\} \:\\\\\\ \qquad:\implies \sf \Big\{ 4 \:+\: 4 \: \Big\} + \Big\{ 16\:-\:15\:\Big\} \:\\\\\\ \qquad:\implies \sf 8 \: + 1\: \:\\\\\\ \qquad:\implies \underline {\boxed {\pmb{\frak{\red{ \: x + y + x y \:=\:9 \:}}}}}\:\:\bigstar \:$

$\qquad \therefore \:\underline {\sf Hence ,\:The \:Value \:of \:x + y + x y\:is\:\pmb{\bf 9\:}\:.}$