|x-2|+3x=5 |x-4|+2x=6 |x|=|-2 1/3| |x-3|-2=4,7 |x-4 1/3|+3 1/4=6 1/4

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1. |x-3|-2=4,7

   |x-3|    = 4,7+2

   |x-3|     =6,7

   TH1: x-3=6,7                                                  TH2: x-3=-6,7

           x    =6,7+3                                                      x   =-6,7+3

           x     =9,7                                                          x   =-3,7

    

   => x thuộc 9,7;-3,7

   Giải thích các bước giải:

    

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2. +, |x – 2| + 3x = 5

   ⇔ |x – 2| = 5 – 3x

   ⇔ \(\left[ \begin{array}{l}x-2=5-3x\\x-2=3x-5\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x+3x=5+2\\x-3x=-5+2\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}4x=7\\-2x=-3\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=\frac{7}{4}\\x=\frac{3}{2}\end{array} \right.\)

   +, |x – 4| + 2x = 6

   ⇔ |x – 4| = 6 – 2x

   ⇔ \(\left[ \begin{array}{l}x-4=6-2x\\x-4=2x-6\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x+2x=6+4\\x-2x=-6+4\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}3x=10\\-x=-2\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=\frac{10}{3}\\x=2\end{array} \right.\)

   +, |x| = |-2$\frac{1}{3}$|

   ⇔ \(\left[ \begin{array}{l}x=2\frac{1}{3}\\x=-2\frac{1}{3}\end{array} \right.\)

   +, |x – 3| – 2 = 4,7

   ⇔ |x – 3| = 4,7 + 2

   ⇔ |x – 3| = 6,7

   ⇔ \(\left[ \begin{array}{l}x-3=6,7\\x-3=-6,7\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=9,7\\x=-3,7\end{array} \right.\)

   +, |x – 4$\frac{1}{3}$|+ 3$\frac{1}{4}$ = 6$\frac{1}{4}$

   ⇔ |x – 4$\frac{1}{3}$| = 3

   ⇔ \(\left[ \begin{array}{l}x-4\frac{1}{3}=3\\x-4\frac{1}{3}=-3\end{array} \right.\)

   ⇔ \(\left[ \begin{array}{l}x=\frac{22}{3}\\x=\frac{4}{3}\end{array} \right.\)

    

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