Question

Bài 2 : Tìm x, biết 

a) \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)                             b) \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)                     c) (5x -1).(2x-\(\frac{1}{3}\)) = 0

Answer 1

\(a.\)

\(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

\(\Rightarrow\left(\frac{3}{5}+x\right)=\frac{3}{35}-\frac{2}{7}=\frac{3}{35}-\frac{10}{35}=-\frac{7}{35}=-\frac{1}{5}\)

\(\Rightarrow x=-\frac{1}{5}-\frac{3}{5}=-\frac{4}{5}\)

Vậy : \(x=-\frac{4}{5}\)

\(b.\)

\(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)

\(\Rightarrow\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}=\frac{3}{14}-\frac{6}{14}=-\frac{3}{14}\)

\(\Rightarrow x=\frac{1}{7}:-\frac{3}{14}=\frac{1}{7}:-\frac{14}{3}=-\frac{2}{3}\)

Vậy : \(x=-\frac{2}{3}\)

\(c.\)

\(\left(5x-1\right).\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}\left(5x-1\right)=0\\\left(2x-\frac{1}{3}\right)=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}5x=1\left(lo\text{ại}\right)\\2x=\frac{1}{3}\end{array}\right.\)

\(\Rightarrow x=\frac{1}{3}:2=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}\)

Vậy : \(x=\frac{1}{6}\)

 

Answer 2

a) \(\frac{2}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\\ =>\frac{3}{5}+x=-\frac{8}{35}\\ =>x=-\frac{29}{35}\)

b) \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\\ =>\frac{1}{7}:x=-\frac{3}{14}\\ =>x=-\frac{28}{21}\\ =>x=-\frac{4}{3}\)

c) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\\ =>5x-1=0v\text{à}2x-\frac{1}{3}=0\).

Nếu \(5x-1=0\\ =>5x=1\left(lo\text{ại}\right)\)

Nếu : \(2x-\frac{1}{3}=0\\ =>2x=\frac{1}{3}\\ =>x=\frac{1}{6}\)

Vậy x=\(\frac{1}{6}\)

Answer 3

a) \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)

=>    \(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)

=>     \(\frac{3}{5}+x=\frac{-1}{5}\)

=>     \(x=\frac{-1}{5}-\frac{3}{5}\) 

=>     \(x=\frac{-4}{5}\)

Vậy \(x=\frac{-4}{5}\)

b)   \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)

=>    \(\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}\)

=>     \(\frac{1}{7}:x=\frac{-3}{14}\)

=>      \(x=\frac{1}{7}:\frac{-3}{14}\)

=>      \(x=\frac{-2}{3}\)

   Vâỵ \(x=\frac{-2}{3}\)

c)  \(\left(5x-1\right).\left(2x-\frac{1}{3}\right)=0\)

=>   \(5x-1=0và2x-\frac{1}{3}=0\)

=>   \(5x=0+1và2x=0+\frac{1}{3}\)

=>   \(5x=1và2x=\frac{1}{3}\)

=>   \(x=1:5vàx=\frac{1}{3}:2\)

=>   \(x=\frac{1}{5}vàx=\frac{1}{6}\)

  Vậy \(x=\frac{1}{5}vàx=\frac{1}{6}\)

 

 

 

 

Answer 4

nhầm 1 tí câu c) chỉ có \(\frac{1}{6}\) là đúng thôi