Matemática, perguntado por mariauiza426, 11 meses atrás


1)calcule \: as \: pontencias \: de \: expoente \: negativo \\  \\ a)7 { }^{ - 1}   \\  \\ b)( \frac{1}{5}) {}^{ - 2} \\  \\ c)( - 0.5) {}^{4} \\  \\ d)( \frac{5}{9}) ^{ - 1} \\  \\ e)( -  \frac{3}{8}) {}^{ - 1} \\  \\ f)( - 3) {}^{ - 3} \\  \\ g)( -  \frac{3}{2}) {}^{ - 2} \\  \\ h)( \frac{7}{4}) {}^{ - 2} \\  \\ i)10 {}^{ - 2} \\  \\ j)( - 1)  {}^{ - 5} \\  \\ k)( \frac{1}{100}) {}^{ - 3} \\  \\ l) - ( - 0.1) {}^{4}
me ajudam pfv com cálculo pfv​

Soluções para a tarefa

Respondido por GeBEfte
2

a)7 { }^{ - 1}~=~\boxed{\frac{1}{7}} \\ \\ b)( \frac{1}{5}) {}^{ - 2}~=~(5)^2~=~\boxed{25}\\ \\ c)( - 0.5) {}^{4}~=~(-\frac{1}{2})^4~=~(\frac{(-1)^4}{2^4})~=~\boxed{\frac{1}{16}} \\ \\ d)( \frac{5}{9}) ^{ - 1}~=~\boxed{\frac{9}{5}} \\ \\ e)( - \frac{3}{8}) {}^{ - 1}~=~\boxed{-\frac{8}{3}} \\ \\ f)( - 3) {}^{ - 3}~=~(-\frac{1}{3})^3~=~(\frac{(-1)^3}{3^3})~=~\frac{-1}{27}~=~\boxed{-\frac{1}{27}}

g)( - \frac{3}{2}) {}^{ - 2}~=~(-\frac{2}{3})^2~=~\frac{(-2)^2}{3^2}~=~\boxed{\frac{4}{9}} \\ \\ h)( \frac{7}{4}) {}^{ - 2}~=~(\frac{4}{7})^2~=~\frac{4^2}{7^2}~=~\boxed{\frac{16}{49}} \\ \\ i)10 {}^{ - 2}~=~(\frac{1}{10})^2~=~\frac{1^2}{10^2}~=~\boxed{\frac{1}{100}} \\ \\ j)( - 1) {}^{ - 5}~=~(-\frac{1}{1})^5~=~\frac{(-1)^5}{1^5}~=~\frac{-1}{1}~=~\boxed{-1} \\ \\ k)( \frac{1}{100}) {}^{ - 3}~=~(\frac{100}{1})^3~=~100^3~=~\boxed{1000000}

l) - ( - 0.1) {}^{4}~=~-(-\frac{1}{10})^4~=~-(\frac{(-1)^4}{10^4})~=~-(\frac{1}{10000})~=~\boxed{-0,0001}


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