câu1
(3x-1).(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
=>x=1/3 =>x=0
câu2
1/4+1/3 :(2x-1)=5
=> 1/3:(2x-1)=19/4
=>2x-1 =57/4
=>2x=61/4
=>x=61/8
còn hai câu sau bn ghi đề mik ko hỉu
1.
a)(3x-1)(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
3x=0+1 x=0:1/2:5
x=1/3 x=0
Vậy x=1/3 hoặc x=0
b)1/4+1/3:(2x-1)=5
1/3:(2x-1)=5-1/4=20/4-1/4=19/4
2x-1=1/3:19/4=1/3*4/19=4/57
2x=4/57+1=4/57+57/57=61/57
x=61/57:2=61/57*1/2=61/114
Vậy x=61/114
c)(2x+2/5)2-9/25=0=02-9/25
=>2x+2/5=0
2x=0-2/5
x=-2/5:2=-2/5*1/2
x=-1/5
Vậy x=-1/5
d)(3x-1/2)3+1/9=0=03+1/9
=>3x-1/2=0
3x=0+1/2
x=1/2:3=1/2*1/3
x=1/6
Vậy x=1/6
Làm nhiêu đó chết người đó bạn 
, đề nhiều số nữa
a) \(\left(3x-1\right)\cdot\dfrac{1}{2\cdot5}=0\)
Vì \(\dfrac{1}{2\cdot5}=\dfrac{1}{10}< 0\)
\(\Rightarrow3x-1=0\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\dfrac{1}{3}\)
Vậy x=\(\dfrac{1}{3}\)
b) \(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=5\)
\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=\dfrac{19}{4}\)
\(\Rightarrow2x-1=\dfrac{4}{57}\)
\(\Rightarrow2x=\dfrac{61}{57}\)
\(\Rightarrow x=\dfrac{61}{114}\)
c) \(\left(2x+\dfrac{2}{5}\right)^2-\dfrac{9}{25}=0\)
\(\Rightarrow\left(2x+\dfrac{2}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-\dfrac{2}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x-\dfrac{2}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{2}{5}=\dfrac{3}{5}\\2x-\dfrac{2}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x=\dfrac{-1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{10}\end{matrix}\right.\)
d) \(\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(\Rightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)
Vì không có trường hợp nào xảy ra để:
\(\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\)
Nên không tồn tại giá trị x thỏa mãn phương trình
Vậy \(x\in\varnothing\)
