a: \(\left(-\dfrac{3}{4}\right)^{3x-1}=-\dfrac{27}{64}\)
=>\(\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^3\)
=>3x-1=3
=>3x=4
=>\(x=\dfrac{4}{3}\)
b: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
=>\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
=>2x+5=4
=>2x=-1
=>\(x=-\dfrac{1}{2}\)
c: \(3,8:2x=\left(\dfrac{1}{2}\right)^2:2\dfrac{2}{3}\)
=>\(3,8:2x=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{1}{4}\cdot\dfrac{3}{8}=\dfrac{3}{32}\)
=>\(2x=3,8:\dfrac{3}{32}=\dfrac{608}{15}\)
=>\(x=\dfrac{304}{15}\)
d: \(\left(x-\dfrac{2}{15}\right)^3=\dfrac{8}{125}\)
=>\(\left(x-\dfrac{2}{15}\right)^3=\left(\dfrac{2}{5}\right)^3\)
=>\(x-\dfrac{2}{15}=\dfrac{2}{5}\)
=>\(x=\dfrac{2}{15}+\dfrac{2}{5}=\dfrac{8}{15}\)
e: \(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
=>\(\dfrac{4}{13}\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
=>\(\dfrac{32}{65}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
=>\(\left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}=\dfrac{10}{13}\cdot\dfrac{65}{32}=5\cdot\dfrac{5}{16}=\dfrac{25}{16}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{4}\\2x=-\dfrac{9}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=-\dfrac{9}{8}\end{matrix}\right.\)
f: ĐKXĐ: x<>-5
\(\dfrac{x-1}{x+5}=\dfrac{5}{6}\)
=>6(x-1)=5(x+5)
=>6x-6=5x+25
=>6x-5x=25+6
=>x=31(nhận)