Question

can anyone solve this??? the right answer get 50 points and will be brainlist and rubbish answer will be reported.​

Answer 1

$\huge\mathcal{\mid{\mid{\underline{\pink{Good\: Night\:}}}{\mid{\mid}}}}$

Qᴜᴇsᴛɪᴏɴ ;-

$\red\bigstar\:\bf\blue{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n}\:\div\:\Big(\dfrac{1}{3}\Big)^{-2}\:\times\:\Big[\dfrac{1}{3^{n/2}}\Big]^{-2}\:-\:\Big[\dfrac{1}{3^{-3}}\Big]^n}{\Big[\dfrac{1}{3^{-2/3}}\Big]^{\frac{9}{2}m}\:\times\:\Big[\dfrac{1}{16^{-3}}\Big]^{\frac{1}{4}}\:}}$

$\pink\bullet\:\bf{if\:m\:-\:n\:=\:(2^{m\:+\:n})^{m\:-\:n}.(2^{n\:+\:p})^{n\:-\:p}.(2^{p\:+\:m})^{p\:-\:m}}$

${\orange{\boxed{\fcolorbox{lime}{indigo}{\color{aqua}ANSWER\:With\: EXPLANATION}}}}$

$\red\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n}\:\div\:\Big(\dfrac{1}{3}\Big)^{-2}\:\times\:\Big[\dfrac{1}{3^{n/2}}\Big]^{-2}\:-\:\Big[\dfrac{1}{3^{-3}}\Big]^n}{\Big[\dfrac{1}{3^{-2/3}}\Big]^{\frac{9}{2}m}\:\times\:\Big[\dfrac{1}{16^{-3}}\Big]^{\frac{1}{4}}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n}\:\div\:\Big(\dfrac{1}{3}\Big)^{-2}\:\times\:\Big[\dfrac{1}{3}\Big]^{\frac{n}{2}\:\times{-2}}\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{\frac{-2}{3}\times\:\frac{9}{2}m}\:\times\:\Big({(16)^3}\Big)^{\frac{1}{4}}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n}\:\div\:\Big(\dfrac{1}{3}\Big)^{-2}\:\times\:\Big[\dfrac{1}{3}\Big]^{-n}\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{16}^{3\times\:\frac{1}{4}}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n\:-\:(-2)\:+\:(-n)}\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:\Big({(16)^{\frac{1}{4}}}\Big)^{3}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-2n\:-\:n\:+\:2}\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{2}^{3}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n\:+\:2}\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{8}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n}\:\times\:\Big(\dfrac{1}{3}\Big)^2\:-\:\Big[\dfrac{1}{3}\Big]^{-3n}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{8}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n}\:\times\:\Big\{\:\Big(\dfrac{1}{3}\Big)^2\:-\:1\:\Big\}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{8}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n}\:\times\:\Big\{\:\dfrac{1}{9}\:-\:1\:\Big\}}{\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:{8}\:}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n}\:\times\:\Big[\dfrac{1}{3}\Big]^{-3m}\:\times\:\Big\{\:\dfrac{1\:-\:9}{9}\:\Big\}}{8}}$

$\longmapsto\:\bf{\dfrac{\Big(\dfrac{1}{3}\Big)^{-3n\:+\:3m}\:\times\:\Big\{\:\dfrac{-8}{9}\:\Big\}}{8}}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\:(m\:-\:n)}\:\times\:\dfrac{-8}{9}\:\times{\dfrac{1}{8}}}$

➝ Now putting the value of (m - n) in the above equation, we get

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\:\Big\{\:(2^{m\:+\:n})^{m\:-\:n}.(2^{n\:+\:p})^{n\:-\:p}.(2^{p\:+\:m})^{p\:-\:m}\Big\}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\:\Big\{\:(2)^{{(m\:+\:n)}\times{(m\:-\:n)}}.(2)^{{(n\:+\:p)}\times{(n\:-\:p)}}.(2)^{{(p\:+\:m)}\times{(p\:-\:m)}}\Big\}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\:\Big\{\:(2)^{m^2\:-\:n^2}.(2)^{n^2\:-\:p^2}.(2)^{p^2\:-\:m^2}\Big\}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\:\Big\{\:(2)^{(m^2\:-\:n^2\:+\:n^2\:-\:p^2\:+\:p^2\:-\:m^2)}\Big\}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\times\:(2)^{0}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\times\:1}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf{\dfrac{1}{27}\:\times\:\Big(\dfrac{-1}{9}\Big)\:}$

$\longmapsto\:\bf\green{-\:\dfrac{1}{243}\:}$

Answer 2

Answer:

$⟼ </p><p>[ </p><p>3 </p><p>−2/3</p><p> </p><p>1</p><p> </p><p> ] </p><p>2</p><p>9</p><p> </p><p> m</p><p> ×[ </p><p>16 </p><p>−3</p><p> </p><p>1</p><p> </p><p> ] </p><p>4</p><p>1$

$( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2n</p><p> ÷( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2</p><p> ×[ </p><p>3 </p><p>n/2</p><p> </p><p>1</p><p> </p><p> ] </p><p>−2</p><p> −[ </p><p>3 </p><p>−3</p><p> </p><p>1</p><p> </p><p> ] </p><p>n</p><p>$

$⟼ </p><p>[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>3</p><p>−2</p><p> </p><p> × </p><p>2</p><p>9</p><p> </p><p> m</p><p> ×((16) </p><p>3</p><p> ) </p><p>4</p><p>1</p><p> </p><p> </p><p> </p><p>( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2n</p><p> ÷( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2</p><p> ×[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>2</p><p>n</p><p> </p><p> ×−2</p><p> −[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3n$

$⟼ </p><p>[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3m</p><p> ×16 </p><p>3× </p><p>4</p><p>1</p><p> </p><p> </p><p> </p><p>( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2n</p><p> ÷( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−2</p><p> ×[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−n</p><p> −[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3n</p><p>$

$⟼ </p><p>[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3m</p><p> ×((16) </p><p>4</p><p>1</p><p> </p><p> </p><p> ) </p><p>3</p><p>$

$⟼ </p><p>[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3m</p><p> ×8</p><p>( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−3n+2</p><p> −[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3n</p><p>$

$⟼ </p><p>8</p><p>( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−3n</p><p> ×[ </p><p>3</p><p>1</p><p> </p><p> ] </p><p>−3m</p><p> ×{ </p><p>9</p><p>1−9</p><p> </p><p> }$

$⟼ </p><p>8</p><p>( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>−3n+3m</p><p> ×{ </p><p>9</p><p>−8</p><p> </p><p> }$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3(m−n)</p><p> × </p><p>9</p><p>−8</p><p> </p><p> × </p><p>8</p><p>1</p><p>$

➝ Now putting the value of (m - n) in the above equation, we get

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3{(2 </p><p>m+n</p><p> ) </p><p>m−n</p><p> .(2 </p><p>n+p</p><p> ) </p><p>n−p</p><p> .(2 </p><p>p+m</p><p> ) </p><p>p−m</p><p> }$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3{(2) </p><p>(m+n)×(m−n)</p><p> .(2) </p><p>(n+p)×(n−p)</p><p> .(2) </p><p>(p+m)×(p−m)</p><p> }</p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p>$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3{(2) </p><p>m </p><p>2</p><p> −n </p><p>2</p><p> </p><p> .(2) </p><p>n </p><p>2</p><p> −p </p><p>2</p><p> </p><p> .(2) </p><p>p </p><p>2</p><p> −m </p><p>2</p><p> </p><p> }</p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p>$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3{(2) </p><p>(m </p><p>2</p><p> −n </p><p>2</p><p> +n </p><p>2</p><p> −p </p><p>2</p><p> +p </p><p>2</p><p> −m </p><p>2</p><p> )</p><p> }</p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )$

$\begin{gathered}\longmapsto\:\bf{\Big(\dfrac{1}{3}\Big)^{3\times\:(2)^{0}}\:\times\:\Big(\dfrac{-1}{9}\Big)\:} \\ \\ \end{gathered}</p><p>$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3×(2) </p><p>0</p><p> </p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p> </p><p> </p><p>$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3×1</p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p> </p><p>$

$⟼( </p><p>3</p><p>1</p><p> </p><p> ) </p><p>3</p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p>$

$⟼ </p><p>27</p><p>1</p><p> </p><p> ×( </p><p>9</p><p>−1</p><p> </p><p> )</p><p> </p><p>$

$➪</p><p>243</p><p>1</p><p> </p><p>$