Question

please answer the questions..................​

Answer 1

Answer:

you can chech that on google

Answer 2

Answer:

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: AnSwer :- \: \star}}}}}$

To prove :--

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

Let's take LHS and prove it to be equal to RHS

$= \frac{ {2}^{36} + ( \frac{1}{4} \times {2}^{35} ) + ( \frac{1}{8} \times {2}^{37} )}{ (\frac{1}{16} \times {2}^{39} ) + (\frac{1}{8} \times {2}^{38} )} \\ \\ = \frac{ {2}^{36} + ( \frac{1}{ {2}^{2} } \times {2}^{35} ) + ( \frac{1}{ {2}^{3} } \times {2}^{37} )}{ (\frac{1}{ {2}^{4} } \times {2}^{39} ) + (\frac{1}{ {2}^{3} } \times {2}^{38} )} \\ \\ = \frac{ {2}^{36} + ( {2}^{35 - 2} ) + ( {2}^{37 - 3} )}{ ( {2}^{39 - 4} ) + ( {2}^{38 - 3} )} \\ \\ = \frac{ {2}^{36} + ( {2}^{33} ) + ( {2}^{34} )}{ ( {2}^{35} ) + ( {2}^{35} )} \\ \\ = \frac{ {2}^{33} ( {2}^{3} + 1 + {2}) }{ {2}^{33}( {2}^{2} + {2}^{2}) }$

Here 2^33 cancels because it is common on both numerator and denominator

$= \frac{2 ^{3} + 3}{4 + 4} \\ \\ = \frac{8 + 3}{8} \\ \\ = \frac{11}{8}$

Hence

$\mathfrak{\huge{\pink{\underline{\underline{proved}}}}}$