\(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)\(\left(n\ge1\rig...">
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20 tháng 10 2021

\(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)

\(=\frac{\left(-7\right)^n}{\left(-7\right)^n:\left(-7\right)}\)

\(=\frac{\left(-7\right)^n}{\left(-7\right)^n.\frac{1}{\left(-7\right)}}\)

\(=\frac{1}{\frac{1}{-7}}\)

\(=-7\)

20 tháng 10 2021

-7 nha bạn

13 tháng 12 2015

\(\frac{\left(-\frac{5}{7}\right)^{n+1}}{\left(-\frac{5}{7}\right)^n}=\frac{\left(-\frac{5}{7}\right)^n.\left(-\frac{5}{7}\right)}{\left(-\frac{5}{7}\right)^n}=\frac{-\frac{5}{7}}{1}=-\frac{5}{7}\)

10 tháng 12 2015

\(\frac{-5}{7}\)

10 tháng 10 2020

a) Ta có: \(\left(0.25\right)^4\cdot1024\)

\(=\left(0.25\right)^4\cdot4^4\cdot4\)

\(=\left(0.25\cdot4\right)^2\cdot4\)

\(=1^2\cdot4=4\)

b) Ta có: \(\frac{230^3}{23^3}\)

\(=\left(\frac{230}{23}\right)^3\)

\(=10^3=1000\)

c) Ta có: \(\frac{\left(-7\right)^n}{\left(-7\right)^{n-1}}\)

\(=\left(-7\right)^n:\left[\frac{\left(-7\right)^n}{-7}\right]\)

\(=\left(-7\right)^n\cdot\frac{-7}{\left(-7\right)^n}\)

\(=-7\)

23 tháng 7 2018

a) \(\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^{n-1}}\)

\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n:\left(-\dfrac{5}{7}\right)}\)

\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{7}{5}\right)}\)

\(=\dfrac{1}{\left(-\dfrac{7}{5}\right)}\)

\(=1.\left(-\dfrac{5}{7}\right)\)

\(=-\dfrac{5}{7}\)

b) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)

\(=\left(-\dfrac{1}{2}\right)^n\)

22 tháng 7 2018

a/ \(\left(2^2\right)^{\left(2^2\right)}=4^4=256\)

b/ \(\dfrac{\left(-\dfrac{5}{7}\right)^{n+1}}{\left(-\dfrac{5}{7}\right)^n}=\dfrac{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{5}{7}\right)}{\left(-\dfrac{5}{7}\right)^n}=-\dfrac{5}{7}\)

c/ \(\dfrac{8^{14}}{4^{12}}=\dfrac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}=\dfrac{2^{42}}{2^{24}}=2^{18}\)

22 tháng 7 2018

thank you

30 tháng 9 2016

a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)

\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)

\(\Leftrightarrow3n-1=2\)

\(\Leftrightarrow3n=3\)

\(\Leftrightarrow n=1\)

b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)

\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)

\(\Leftrightarrow n+2=-1\)

\(\Leftrightarrow n=-3\)

c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)

\(\Leftrightarrow-n+1=-3\)

\(\Leftrightarrow n=-4\)

c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)

\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)

\(\Leftrightarrow3n+1=-3\)

\(\Leftrightarrow3n=-4\)

\(\Leftrightarrow n=-\frac{4}{3}\)

a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)

b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)