Question

Ayuda por favor es examen

Answer 1

Explicación paso a paso:

Tarea:

• Resolver.

Solución:

Recordar:

• En una multiplicación de bases iguales el exponente se suma.
• En una división de bases iguales los exponentes se restan.
• ${12}^{2} \times {12}^{2} \times {12}^{6} = {12}^{10}$
• ${ \frac{4}{2} }^{3} \times { \frac{4}{2} }^{5} \times { \frac{4}{2} }^{4} = { \frac{4}{2} }^{12}$
• ${15}^{2} \times {15}^{4} \times {15}^{ \frac{1}{2} } = {15}^{6 + \frac{1}{2} } = {15}^{ \frac{13}{2} }$
• ${2}^{ \frac{1}{3} } \times {2}^{ \frac{1}{3} } \times {2}^{ \frac{1}{3} } = {2}^{ \frac{3}{3} } = {2}^{1} = 2$
• $( \frac{5}{8} ) {}^{ \frac{4}{6} } \times (\frac{5}{8} ) {}^{ \frac{4}{6} } \times (\frac{5}{8} ) {}^{ \frac{4}{6} } =( \frac{5}{8} ) {}^{ \frac{12}{6} } = ( \frac{5}{8} ) {}^{2}$
• $\frac{ {2}^{8} }{ {2}^{4} } = {2}^{8 - 4} = {2}^{2}$
• $\frac{ {15}^{21} }{ {15}^{12} } = {15}^{21 - 12} = {15}^{9}$
• $\frac{ {12}^{7} }{ {12}^{11} } = {12}^{7 - 11} = {12}^{ - 4}$
• $\frac{ {16}^{ - 8} }{ {16}^{ - 2} } = {16}^{8 - ( - 2)} = {16}^{8 + 2} = {16}^{10}$
• $({5}^{2} ) \frac{8}{4} = {5}^{ \frac{16}{4} } = {5}^{4}$
• $({17}^{5}) \frac{40}{4} = ( {17}^{2} ) {}^{10} = {17}^{20}$
• $(( \frac{2}{5} ) {}^{2} ) {}^{2} = ({ \frac{2}{5} )}^{4}$
• $( {64}^{2} ) {}^{ - 2} = {64}^{ - 4}$
• $(( \frac{9}{5} ) {}^{ - 3} ) {}^{ - 2} = ( \frac{9}{5} ) {}^{6}$
• Espero le sirva.