Question

a) $(\frac{1}{3})^5. (\frac{1}{3}). (\frac{1}{3})^8: (\frac{1}{3})^9$ 

b) $[-(-\frac{1}{2})^{2}]^{-3}: [(-\frac{1}{2}).(-\frac{1}{2})^{5}.(-\frac{1}{2})^{-3}]$

c) $\left \{ {{2x+y=12} \atop {x+3y=11}} \right.$

d) $\left \{ {{x=5y} \atop {x+y=12}} \right.$

e) $\frac{1}{2}. \frac{2}{3}. \frac{3}{4}. \frac{4}{5}. \frac{5}{6}. \frac{6}{7}. \frac{7}{8}. \frac{8}{9}. \frac{9}{10}$




Ficarei muito grato!

Answer 1

$a) \\ \\ ( \dfrac{1}{3} )^5.( \dfrac{1}{3} ).( \dfrac{1}{3} )^8:( \dfrac{1}{3} )^9 \\ \\ ( \dfrac{1}{3} )^{(5+1+8-9)} \\ \\ ( \dfrac{1}{3} )^{5} \\ \\ ( \dfrac{1}{3^5} )\\ \\ ( \dfrac{1}{243} )$

$b) \\ \\ |-(-\dfrac{1}{2})^2]^{-3} : [( -\dfrac{1}{2}) . (-\dfrac{1}{2})^5. (- \dfrac{1}{2} )^{-3} ] \\ \\ (-\dfrac{1}{4})^{-3} : [( -\dfrac{1}{2})^{1+5-3}] \\ \\ (-\dfrac{4}{1})^{3} : [( -\dfrac{1}{2})^3] \\ \\ (-4)^{3} : ( -\dfrac{1}{8}) \\ \\ -64 : ( -\dfrac{1}{8}) \\ \\ -64\ .\ (-8)\\ \\ 512$

$c) \\  \\ \left \{ {{2x+y=12\ \ \ \ \ \ \ \ } \atop {x+3y=11}\ \ .(-2)} \right. \\ \\ \left \{ {{2x+y=12\ \ \ \ } \atop {-2x-6y=-22}} \right. \\ \\ Agora\ somamos\ as\ equa\c{c}\~oes:\\ \\ 0x-5y=-10\\ \\ y = \frac{-10}{-5}\\ \\ y = 2\\ \\ \\ 2x+y=12\\ 2x + 2 = 12\\ 2x = 10\\ x = 5$

$d) \\ \\ \left \{ {{x=5y} \atop {x+y=12}} \right. \\ \\ x+y=12\\ 5y + y = 12\\ 6y = 12\\ y = 2\\ \\ x = 5y\\ x = 5.2\\ x = 10$

$e) \\ \\ \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . \frac{5}{6} . \frac{6}{7} . \frac{7}{8} . \frac{8}{9} . \frac{9}{10}$

Aqui só simplicaremos numerador com denominador (2 com 2, 3 com 3, 4 com 4...)

$\frac{1}{1} . \frac{1}{1} . \frac{1}{1} .\frac{1}{1}. \frac{1}{1} .\frac{1}{1} .\frac{1}{1} .\frac{1}{1} .\frac{1}{10} \\ \\ \frac{1}{10}$

=)