i) \(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\\dfrac{-1}{2}x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=10\)
j) \(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\Rightarrow\dfrac{1}{3}.\left(\dfrac{1}{2x-1}\right)=-5-\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{3}\left(\dfrac{1}{2x-1}\right)=\dfrac{-21}{4}\)
\(\Rightarrow\dfrac{1}{2x-1}=\dfrac{-63}{4}\)
\(\Rightarrow\left(-63\right)\left(2x-1\right)=4\)
\(\Rightarrow-126x+63=4\)
\(\Rightarrow-126x=-59\)
\(\Rightarrow x=\dfrac{59}{126}\)
Vậy \(x=\dfrac{59}{126}\)
i)để (3x-1)(-1/2x+5)=0
=>3x-1=0 hoặc -1/2x+5=0
với 3x-1=0 với -1/2x+5=0
=> 3x=1 => -1/2x=5
=>x=1/3 =>x=5:(-1/2)
=>x=-10
j) 1/4+1/3:(2x-1)=-5
=> 1/3:(2x-1)=-5 - 1/4
=>1/3:(2x-1)=-21/4
=>2x-1=1/3:(-21/4)
=>2x-1=-4/63
=>2x=-4/63+1
=>2x=59/63
=>x=59/63:2
=>x=59/126
like cho mình nhé
