ta có : 

\(\frac{x+1}{3}+\frac{x+1}{9}+\frac{x+1}{27}+\frac{x+1}{81}=\left(x+1\right)\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)

\(=\left(x+1\right)\times\frac{40}{81}=\frac{56}{81}\text{ nên }x+1=\frac{56}{40}=\frac{7}{5}\)

vậy \(x=\frac{7}{5}-1=\frac{2}{5}\)

16/81

a) \(3^{a+1}=81\)

\(3^{a+1}=3^4\)

\(a+1=4\)

\(a=3\)

b) \(\left(x-1\right)^3=27\)

\(\left(x-1\right)^3=3^3\)

\(x-1=3\)

\(x=4\)

\(a,3^{a+1}=81\\ \Rightarrow3^{a+1}=3^4\\ \\ \Rightarrow a+1=4\\ \Rightarrow a=3.\\ b,\left(x-1\right)^3=27\\ \Rightarrow\left(x-1\right)^3=3^3\\ \Rightarrow x-1=3\\ \Rightarrow x=4.\)

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

\(\Leftrightarrow x=\frac{1}{-3}\)

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{6}\)

\(x+\dfrac{40}{27}=2\)
\(x=\dfrac{14}{27}\)

\(x+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}=2\)

\(\Leftrightarrow x+\dfrac{121}{81}=2\)

hay \(x=\dfrac{41}{81}\)

a) \(\left(\frac{-3}{4}\right)^{2n-1}=\frac{81}{256}\Leftrightarrow\left(\frac{-3}{4}\right)^{2n-1}=\left(\frac{-3}{4}\right)^4\)

<=>2n-1=4

<=>2n=5

<=>n=5/2

b)\(27

a: x=2/5*15=6

b: =>x-7=15

=>x=22

c: =>3/4:x+1/2*4=4

=>3/4:x=4-2=2

=>x=3/8

không biết