Question

$\frac{2^{x+3} + 2^{x+2}- 2^{x-1}}{2^{x-2}+2^{x}}$

Answer 1

Resposta:

23/3

Explicação passo-a-passo:

$\frac{2^{x+3} + 2^{x+2}- 2^{x-1}}{2^{x-2}+2^{x}} \\ \\ \frac{ {2}^{x} . {2}^{3} + {2}^{x}. {2}^{2} - {2}^{x}. {2}^{ - 1} }{ {2}^{x}. {2}^{ - 2} + {2}^{x}} \\ \\ \frac{ {2}^{x}.( {2}^{3} + {2}^{2} - {2}^{ - 1})}{ {2}^{x}.( {2}^{ - 2} + 1)} \\ \\ \frac{ {2}^{3} + {2}^{2} - {2}^{ - 1} }{ {2}^{ - 2} + 1} \\ \\ \frac{8 + 4 - \frac{1}{2} }{ \frac{1}{4} + 1} \\ \\ \frac{12 - \frac{1}{2} }{ \frac{5}{4} } \\ \\ \frac{ \frac{23}{2} }{ \frac{5}{4} } \\ \\ \frac{23}{2} \times \frac{4}{5} \\ \\ \frac{23\times 2}{5} \\ \\ \frac{46}{5}$

Answer 2

Resposta:

${46\over5}$

Explicação passo-a-passo:

Desmembrando

${2^x.2^3+2^x.2^2-2^x.2^{-1}\over2^x.2^{-2}+2^x}=\\ \\ fatorar~~com~~2^x~~fator~~comum\\ \\ {2^x(2^3+2^2-2^{-1})\over2^x(2^{-2}+1)}=\\ \\ cancela~~2^x\\ \\ (8+4-{1\over2})\div({1\over4}+1)=\\ \\ ({16+8-1\over2})\div({1+4\over4})=\\ \\ ({23\over2})\div({5\over4})=$

$({23\over\not2})\times({\not4^2\over5})=\fbox{ {46\over5} }$