Question

Prove that :( help me​

Answer 1

${\boxed{ \boxed{ \bold \red{Please \: mark \: as \: brainliest}}}}$

Answer 2

Prove that

$\rm \bigg( {\bigg(8 \bigg) }^{ - \dfrac{2}{3} } \times {\bigg(2\bigg) }^{\dfrac{1}{2} } \times {\bigg(25 \bigg) }^{ - \dfrac{5}{4} } \bigg) \div \bigg({\bigg(32\bigg) }^{ - \dfrac{2}{5} } \times {\bigg(125 \bigg) }^{ - \dfrac{5}{6} }\bigg) = \sqrt{2}$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\bigg( {\bigg(8 \bigg) }^{ - \dfrac{2}{3} } \times {\bigg(2\bigg) }^{\dfrac{1}{2} } \times {\bigg(25 \bigg) }^{ - \dfrac{5}{4} } \bigg) \div \bigg({\bigg(32\bigg) }^{ - \dfrac{2}{5} } \times {\bigg(125 \bigg) }^{ - \dfrac{5}{6} }\bigg)$

Let we first simply each term in its simplest form.

Consider,

$\rm :\longmapsto\:{\bigg(8\bigg) }^{ - \dfrac{2}{3} }$

$\rm \: = \: \:\:{\bigg(2 \times 2 \times 2\bigg) }^{ - \dfrac{2}{3} }$

$\rm \: = \: \:\:{\bigg( {(2)}^{3} \bigg) }^{ - \dfrac{2}{3} }$

$\rm \: = \: \: {(2)}^{ - 2}$

Now, Consider,

$\rm :\longmapsto\:{\bigg( 25\bigg) }^{ - \dfrac{5}{4} }$

$\rm \: = \: \:\:{\bigg( 5 \times 5\bigg) }^{ - \dfrac{5}{4} }$

$\rm \: = \: \:\:{\bigg( \sqrt{5}\times \sqrt{5} \times \sqrt{5} \times \sqrt{5} \bigg) }^{ - \dfrac{5}{4} }$

$\rm \: = \: \:\:{\bigg( {( \sqrt{5}) }^{4} \bigg) }^{ - \dfrac{5}{4} }$

$\rm \: = \: \: {( \sqrt{5} )}^{ - 5}$

Now, Consider,

$\rm :\longmapsto\:{\bigg(32\bigg) }^{ - \dfrac{2}{5} }$

$\rm \: = \: \:\:{\bigg(2 \times 2 \times 2 \times 2 \times 2\bigg) }^{ - \dfrac{2}{5} }$

$\rm \: = \: \:\:{\bigg( {(2)}^{5} \bigg) }^{ - \dfrac{2}{5} }$

$\rm \: = \: \: {(2)}^{ - 2}$

Now Consider,

$\rm :\longmapsto\:{\bigg(125\bigg) }^{ - \dfrac{5}{6} }$

$\rm \: = \: \:\:{\bigg( 5 \times 5 \times 5 \bigg) }^{ - \dfrac{5}{6} }$

$\rm \: = \: \:\:{\bigg( \sqrt{5}\times \sqrt{5} \times \sqrt{5} \times \sqrt{5} \times \sqrt{5} \times \sqrt{5} \bigg) }^{ - \dfrac{5}{6} }$

$\rm \: = \: \:\:{\bigg( {( \sqrt{5}) }^{6} \bigg) }^{ - \dfrac{5}{6} }$

$\rm \: = \: \: {( \sqrt{5} )}^{ - 5}$

Consider given expression,.

$\rm :\longmapsto\:\bigg( {\bigg(8 \bigg) }^{ - \dfrac{2}{3} } \times {\bigg(2\bigg) }^{\dfrac{1}{2} } \times {\bigg(25 \bigg) }^{ - \dfrac{5}{4} } \bigg) \div \bigg({\bigg(32\bigg) }^{ - \dfrac{2}{5} } \times {\bigg(125 \bigg) }^{ - \dfrac{5}{6} }\bigg)$

So, Now given expression can be rewritten as

$\rm \: = \: \:\dfrac{{\bigg(8 \bigg) }^{ - \dfrac{2}{3} } \times {\bigg(2\bigg) }^{\dfrac{1}{2} } \times {\bigg(25 \bigg) }^{ - \dfrac{5}{4} }}{{\bigg(32\bigg) }^{ - \dfrac{2}{5} } \times {\bigg(125 \bigg) }^{ - \dfrac{5}{6} }}$

Now, on substituting the values, evaluated above, we get

$\rm \: = \: \:\dfrac{{\bigg(2 \bigg) }^{ - 2} \times {\bigg(2\bigg) }^{\dfrac{1}{2} } \times {\bigg( \sqrt{5} \bigg) }^{ - 5 }}{{\bigg(2\bigg) }^{ - 2 } \times {\bigg( \sqrt{5} \bigg) }^{ - 5}}$

$\rm \: = \: \:{\bigg(2\bigg) }^{\dfrac{1}{2} }$

$\rm \: = \: \: \sqrt{2}$

$\boxed{\boxed{\bf{Hence, Proved}}}$

Additional Information :-

$\boxed{ \bf{ \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n}}}$

$\boxed{ \bf{ \: {x}^{m} \div {x}^{n} \: = \: {x}^{m - n}}}$

$\boxed{ \bf{ \: ({x}^{m})^{n} \: = \: {x}^{mn}}}$

$\boxed{ \bf{ \: {x}^{0}\: = \: 1}}$