a) \(x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{1}{5}\)
\(x=\dfrac{11}{20}\)
b) \(x+\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{2}{3}\)
\(x+\dfrac{2}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}-\dfrac{2}{3}\)
\(x=0\)
c) \(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{729}\)
\(x=6\)
a, x+\(\dfrac{1}{5}\)=\(\dfrac{3}{4}\)
x = \(\dfrac{3}{4}\)-\(\dfrac{1}{5}\)
x =\(\dfrac{11}{20}\)
vậy x=\(\dfrac{11}{20}\)
b, x+\(\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{2}{3}\)
x =\(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{6}\)
x =0
vậy x=0
c,\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{729}\)
\(\left(\dfrac{1}{3^{ }}\right)^x=\left(\dfrac{1}{3}\right)^6\)
=> x=6
vậy x=6