Question

tìm x , biết

2/7 x-1/3=3/5x-1

|3x+7|-3x=7

c) |2x-1|+|3x+2|=0

Answer 1

vt rõ đề câu a đi lm 1 thể

 

Answer 2

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{3}{5}\)\(x\) - 1

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{3}{5}\)\(x\) =  - 1 + \(\dfrac{1}{3}\)

 (\(\dfrac{2}{7}\) - \(\dfrac{3}{5}\))\(x\) =  - \(\dfrac{2}{3}\)

  - \(\dfrac{11}{35}\)\(x\)     = - \(\dfrac{2}{3}\)

         \(x\)     = - \(\dfrac{2}{3}\) : (- \(\dfrac{11}{35}\))

         \(x\)     = \(\dfrac{70}{33}\)

Vậy \(x=\dfrac{70}{33}\) 

Answer 3

|3\(x\) + 7| - 3\(x\) = 7

|3\(x\) + 7| = 7  +3\(x\) Vì 7 + 3\(x\) > 0  nên 3\(x\) >  -7 ⇒ \(x\) >  -\(\dfrac{7}{3}\)

Với \(x\) >  - \(\dfrac{7}{3}\) ta có:

3\(x\) + 7 = 7 + 3\(x\) 

        7  =7 (luôn đúng)

Vậy \(x\) \(\ge\) - \(\dfrac{7}{3}\) 

 

 

Answer 4

c; |2\(x\) - 1| + |3\(x\) + 2| = 0

    |2\(x\) - 1| ≥ 0; |3\(x\) + 2| ≥ 0

     ⇒ |2\(x\) - 1| + |3\(x\) + 2| = 0

     ⇔ \(\left\{{}\begin{matrix}2x-1=0\\3x+2=0\end{matrix}\right.\)

      ⇒ \(\left\{{}\begin{matrix}2x=1\\3x=-2\end{matrix}\right.\)

      \(\Rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

   Vì \(\dfrac{1}{2}\)  > - \(\dfrac{2}{3}\) Vậy  \(x\) \(\in\) \(\varnothing\)