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

\(\Rightarrow\dfrac{1^m}{3^m}=\dfrac{1}{81}\)

\(\Rightarrow\dfrac{1}{3^m}=\dfrac{1}{3^4}\)

\(\Rightarrow m=4\)

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

\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left[\left(\dfrac{3}{5}\right)^2\right]^5\)

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

\(\Rightarrow n=10\)

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

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

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

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

\(\Rightarrow p=4\)

\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)

\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)

B

B.n=5

B

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 :        \(p=4\)

Lời giải:
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+3}=\frac{9}{256}\)

\(\Leftrightarrow \left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^3=\frac{9}{256}\)

\(\Leftrightarrow \left(\frac{1}{2}\right)^x.(1+\frac{1}{8})=\frac{9}{256}\Rightarrow \left(\frac{1}{2}\right)^x=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)

\(\Rightarrow x=5\)

Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)

a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)

=>2x=-45/22

hay x=-45/44

b: =>x/7=-1/28:1/4=-1/7

=>x=-1

a)(7/2+2x).11/3=16/3

7/2+2x=16/3:11/3

7/2+2x=16/3.3/11

7/2+2x=16/11

2x=16/11-7/2

2x= -45/22

x= -45/22:2

x= -45/44

Vậy x= -45/44

b)x/7 +1/4= -1/28

x/7= -1/28-1/4

x/7= -2/7

=>x= -2

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)

\(\Leftrightarrow\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{637}{1275}\)

\(\Leftrightarrow\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}-\dfrac{637}{1275}=\dfrac{1}{2550}\)

\(\Leftrightarrow\left(n+1\right)\left(n+2\right)=2550\)

\(\Leftrightarrow n^2+3n-2548=0\)

\(\Rightarrow n=49\)

@Nguyễn Việt Lâm @Trần Trung Nguyên