\(a.\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\\ TH1:2x-3=0\\ =>2x=3\\ =>x=\dfrac{3}{2}\\ TH2:\dfrac{3}{4}x+1=0\\ =>\dfrac{3}{4}x=-1\\ =>x=-1:\dfrac{3}{4}\\ =>x=\dfrac{-4}{3}\\ b.\dfrac{3}{7}x+2\dfrac{3}{5}=1\dfrac{2}{5}\\ =>\dfrac{3}{7}x=1\dfrac{2}{5}-2\dfrac{3}{5}\\ =>\dfrac{3}{7}x=-\dfrac{6}{5}\\ =>x=\dfrac{-6}{5}:\dfrac{3}{7}=\dfrac{-14}{5}\\ c.\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\\ TH1:5x-1=0\\ =>5x=1\\ =>x=\dfrac{1}{5}\\ TH2:2x-\dfrac{1}{3}=0\\ =>2x=\dfrac{1}{3}\\ =>x=\dfrac{1}{3}:2=\dfrac{1}{6}\\ \dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ =>\dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}\\ =>\dfrac{1}{7}:x=\dfrac{-3}{14}\\ =>x=\dfrac{1}{7}:\dfrac{-3}{14}=-\dfrac{2}{3}\)
`#3107.101107`
`a)`
\(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy, `x \in {3/2; -4/3}`
`b)`
\(\dfrac{3}{7}x+2\dfrac{3}{5}=1\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{7}x=\dfrac{7}{5}-\dfrac{13}{5}\\ \Rightarrow\dfrac{3}{7}x=-\dfrac{6}{5}\\ \Rightarrow x=-\dfrac{14}{5}\)
Vậy, `x = -14/5`
`c)`
\(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy, `x \in {1/5; 1/6}`
`d)`
\(\dfrac{3}{7}+\dfrac{1}{7}\div x=\dfrac{3}{14}\)
$\Rightarrow $ `3/7 + 1/7 * 1/x = 3/14`
$\Rightarrow $ `1/(7x) = 3/14 - 3/7`
$\Rightarrow $ `1/(7x) = -3/14`
$\Rightarrow $ `-3*7x = 14`
$\Rightarrow $ `-21x = 14`
$\Rightarrow $ `x = -2/3`
Vậy, `x = -2/3.`
