a) Ta có
(2x - 3) \(\left(\dfrac{3}{4}x+1\right)\) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\)
TH1: 2x - 3 = 0
2x = 0 + 3
2x = 3
x = 3:2
x = \(\dfrac{3}{2}\)
TH2: \(\dfrac{3}{4}x+1=0\)
\(\dfrac{3}{4}x\) = 0 - 1
\(\dfrac{3}{4}x\) = -1
x = -1 : \(\dfrac{3}{4}\)
x = \(\dfrac{-4}{3}\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)là giá trị cần tìm
b)(5x - 1) \(\left(2x-\dfrac{1}{3}\right)\) = 0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
Th1: 5x - 1 = 0
5x = 0+1
5x = 1
x = 1 : 5
x = \(\dfrac{1}{5}\)
Th2: \(2x-\dfrac{1}{3}\) = 0
2x = 0 + \(\dfrac{1}{3}\)
2x = \(\dfrac{1}{3}\)
x = \(\dfrac{1}{3}:2\)
x = \(\dfrac{1}{6}\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\) là giá trị cần tìm
c) Ta có
\(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
\(\dfrac{1}{7}+x\) = \(\dfrac{3}{14}:\dfrac{3}{7}\)
\(\dfrac{1}{7}+x\) = \(\dfrac{1}{2}\)
x = \(\dfrac{1}{2}-\dfrac{1}{7}\)
x = \(\dfrac{5}{14}\)
Vậy x = \(\dfrac{5}{14}\) là giá trị cần tìm
