Question

please can anyone solve no 47​

Answer 1

${ \dagger{ \bold{ \red{ \: \: given}}}}$



${ \star{ \bold{ \: \: \: {a}^{2} + {b}^{2} + {c}^{2} = 2abc}}}$




${ \dagger{ \bold{ \red{ \: \: \: simplify}}}}$




${ \bold{ \purple{ {( {2}^{ \frac{1}{ab} }) }^{c} . {( {2}^{ \frac{1}{bc} }) }^{a} . {( {2}^{ \frac{1}{ac} } )}^{b} }}}$




${ \dagger{ \bold{ \red{ \: \: \: solution}}}}$




${ \bold{ = {( {2}^{ \frac{1}{ab} }) }^{c}. {( {2}^{ \frac{1}{bc} }) }^{a} . {( {2}^{ \frac{1}{ac} }) }^{b} }}$





${ \bold{ = ( {2}^{ \frac{c}{ab} } ).( {2}^{ \frac{a}{bc} }). {( {2}^{ \frac{b}{ac} } )}}}$




${ \bold{ = {2}^{( \frac{c}{ab} + \frac{a}{bc} + \frac{b}{ac}) } }}$





${ \bold{ = {2}^{( \frac{ {c}^{2} + {a}^{2} + {b}^{2} }{abc}) }}}$





${ \bold{ = {2}^{( \frac{ {a}^{2} + {b}^{2} + {c}^{2} }{abc}) }}}$





${ \bold{ = {2}^{( \frac{2abc}{abc} )} }}$





${ \bold{ = {2}^{2} }}$




${ \bold{ \huge{ \purple{ = 4}}}}$