\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)

\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)

\(\Rightarrow n+1=4\Rightarrow n=3\)

\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)

\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)

a) Câu này thiếu đề nhé bạn.

b) \(\frac{25}{5^n}=5\)

\(\Rightarrow5^n=25:5\)

\(\Rightarrow5^n=5\)

\(\Rightarrow5^n=5^1\)

\(\Rightarrow n=1\)

Vậy \(n=1.\)

c) \(\frac{81}{\left(-3\right)^n}=-243\)

\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)

\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1.\)

e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)

\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)

\(\Rightarrow n=4\)

Vậy \(n=4.\)

Chúc bạn học tốt!

d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=288:\frac{9}{2}\)

\(\Rightarrow2^n=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

Vậy \(n=6.\)

g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)

\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)

\(\Rightarrow n=3\)

Vậy \(n=3.\)

h) \(5^{-1}.25^n=125\)

\(\Rightarrow5^{-1}.5^{2n}=5^3\)

\(\Rightarrow5^{-1+2n}=5^3\)

\(\Rightarrow-1+2n=3\)

\(\Rightarrow2n=3+1\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=4:2\)

\(\Rightarrow n=2\)

Vậy \(n=2.\)

k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)

\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)

\(\Rightarrow3^{n-1}.7=7.3^6\)

\(\Rightarrow n-1=6\)

\(\Rightarrow n=6+1\)

\(\Rightarrow n=7\)

Vậy \(n=7.\)

Chúc bạn học tốt!

b)\(\frac{25}{5^n}\)=5

\(5^n\)=25/5

\(5^n\)=\(5^1\)

⇒ n = 1

c) \(\frac{81}{\left(-3\right)^n}\)=-243

\(\left(-3\right)^n\)=81/-243

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

\(\left(-3\right)^n\)=\(\left(-3\right)^{-1}\)

⇒n=-1

a) x=1

b) x=1

c) x= -(245/81)

d) x= 1/27

e) x=3

g) x=4

cần gấp

a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)

       \(\Leftrightarrow n=5\)

c ) \(\left(-0,25\right)^p=\frac{1}{256}\)

   \(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)

   \(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)

   \(\Leftrightarrow p=4\)

\(a.\)

\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

Vậy :        \(m=4\)

\(b.\)

\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)

\(\Rightarrow n=5\)

Vậy :        \(n=5\)

\(c.\)

\(\left(-0,25\right)^p=\frac{1}{256}\)

\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)

\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

Vậy :        \(p=4\)

ko bt làm thì xuống lớp 6 hocj đi

Bạn 12345678901 xuống lớp 1 học đạo đức làm người nhé bạn. Lịch sự tí đi

Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)

a/ \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Leftrightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Leftrightarrow m=4\left(tm\right)\)

b/ \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\Leftrightarrow n=10\)

\(\Leftrightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

a)(1/3)^m=(1/3)^4

b)(3/5)^n=(3/5)^10

c)(-0,25)^p=(-0,25)^4

a)\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

 \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)

\(\Rightarrow m=4\)

vay \(m=4\)

b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)

\(\Rightarrow n=10\)

vay \(n=10\)

c) \(\left(-0,25\right)^p=\frac{1}{256}\)

\(\left(\frac{-1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)

\(\Rightarrow p=4\)

voi mu duong thi \(\left(\frac{-1}{4}\right)^4=\left(\frac{1}{4}\right)^4\)

vay \(p=4\)