Question

\(\dfrac{-2}{3}\).(x-\(\dfrac{1}{4}\)) = \(\dfrac{1}{3}\)(2x-1) Tìm x y

\(\dfrac{1}{5}\).\(^{2^x}\)+\(\dfrac{1}{3}\).\(2^{x+1}\)=\(\dfrac{1}{5}\).\(2^7\)+\(\dfrac{1}{3}\).\(2^8\)Tìm xy

Answer 1

\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)

\(-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ -\dfrac{2}{3}x+\dfrac{1}{6}-\dfrac{2}{3}x+\dfrac{1}{3}=0\)

\(-\dfrac{4}{3}x+\dfrac{1}{2}=0\\ -\dfrac{4}{3}x=-\dfrac{1}{2}\\ x=\dfrac{3}{8}\)

\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^{x+1}=\dfrac{1}{5}2^7+\dfrac{1}{3}2^8\)

\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^x\cdot2=\dfrac{1}{5}2^7+\dfrac{1}{3}2^7\cdot2\)

\(2^x\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)

\(2^x=2^7\\ x=7\)