ét ô étttt
giúp tớ...........................
a: =>2x-3=2/5 hoặc 2x-3=-2/5
=>2x=17/5 hoặc 2x=13/5
=>x=17/10 hoặc x=13/10
c: =>2x-1=3
=>2x=4
hay x=2
f: =>x-3=4
hay x=7
\(a,\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ 2x-3=\dfrac{2}{5}\\ 2x=\dfrac{2}{5}+3\\ 2x=\dfrac{2+3.5}{5}\\ 2x=\dfrac{17}{5}\\ x=\dfrac{17}{5}:2\\ x=\dfrac{17}{10}\)
TH2 \(\left(2x-3\right)^2=\left(-\dfrac{2}{5}\right)^2\\ 2x-3=-\dfrac{2}{5}\\ 2x=-\dfrac{2}{5}+3\\ 2x=\dfrac{13}{5}\\ x=\dfrac{13}{10}\)
\(b,\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\\ 2x-1=3\\ 2x=3+1\\ 2x=4\\ x=4:2\\ x=2\\ d,\left(x-\dfrac{5}{6}\right)^2=\left(\dfrac{1}{6}\right)^2\\ x-\dfrac{5}{6}=\dfrac{1}{6}\\ x=\dfrac{1}{6}+\dfrac{5}{6}\\ x=\dfrac{6}{6}\\ x=1\\ TH2\\ \left(x-\dfrac{5}{6}\right)^2=\left(-\dfrac{1}{6}\right)^2\\ x-\dfrac{5}{6}=-\dfrac{1}{6}\\ x=-\dfrac{1}{6}+\dfrac{5}{6}\\ x=\dfrac{2}{3}\\ f,\left(-\dfrac{1}{3}\right)^{x-3}=\left(-\dfrac{1}{3}\right)^4\\ x-3=4\\ x=4+3\\ x=7\)
