Đáp án:

Giải thích các bước giải:

 `a)` `|0,5x-2|-|x+2/3|=0`

`|0,5x-2|=|x+2/3|`

`0,5x-2=x+2/3` hoặc `0,5x-2=-(x+2/3)`

`x=-16/3` hoặc `x=8/9`

Vậy `x` $\in$ `{-16/3;8/9}`

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`b)` `2x-|x+1|=1/4`

`2x-(x+1)=1/4, x+1` $\ge$ `0` hoặc `2x-[-(x+1)] = 1/4, x+1<0`

`x=5/4, x` $\ge$  `-1` hoặc `x=-1/4, x<-1`

`x=5/4` hoặc `x` không có giá trị thỏa mãn`

Vậy `x = 5/4`

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`c)` `3x-|x+15|=5/4`

`3x-(x+15)=5/4,x+15` $\ge$ `0` hoặc `3x-[-(x+15)]=5/4,x+15<0`

`x=65/8, x` $\ge$ `-15` hoặc `x=-55/16, x<-15`

`x=65/8` hoặc `x` không có giá trị thỏa mãn.

Vậy `x=65/8`

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`d)` `3/2-|5/4+3x|=1/4`

`-|5/4+3x|=1/4-3/2`

`-|5/4+3x|=-5/4`

`|5/4+3x|=5/4`

`5/4+3x=5/4` hoặc `5/4+3x=-5/4`

`x=0` hoặc `x=-5/6`

Vậy `x` $\in$ `{-5/6;0}`

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`e)` `|4x-1|=|3x-1/2|`

`4x-1=3x-1/2` hoặc `4x-1=-(3x-1/2)`

`x=1/2` hoặc `x=3/14`

Vậy `x` $\in$ `{3/14;1/2}`

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`f)` `|2x-1|=|x+1/3|`

`2x-1=x+1/3` hoặc `2x-1=-(x+1/3)`

`x=4/3` hoặc `x=2/9`

Vậy `x` $\in$ `{2/9;4/3}`

` a)|0,5x-2|-|x+2/3|=0 `

` =>|1/2 x-2|=|x+2/3| `

` => ` \(\left[ \begin{array}{l}\dfrac{1}{2}x-2=x+\dfrac{2}{3}\\\dfrac{1}{2}x-2=-x-\dfrac{2}{3}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}\dfrac{1}{2}x-x=\dfrac{2}{3}+2\\\dfrac{1}{2}x+x=-\dfrac{2}{3}+2\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}-\dfrac{1}{2}x=\dfrac{8}{3}\\\dfrac{3}{2}x=\dfrac{4}{3}\end{array} \right.\) ` => ` \(\left[ \begin{array}{l}x=-\dfrac{16}{3}\\x=\dfrac{8}{9}\end{array} \right.\) 

` b)2x-|x+1|=1/4 `

` =>|x+1|=2x-1/4 `

` ĐKXĐ:2x-1/4≥0=>2x≥1/4=>x≥1/8 `

` => ` \(\left[ \begin{array}{l}x+1=2x-\dfrac{1}{4}\\x+1=-2x+\dfrac{1}{4}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x-2x=-\dfrac{1}{4}-1\\x+2x=\dfrac{1}{4}-1\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}-x=-\dfrac{5}{4}\\3x=-\dfrac{3}{4}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x=\dfrac{5}{4}(TMĐK)\\x=-\dfrac{1}{4}(KTMĐK)\end{array} \right.\)

` =>x=5/4 `

` c)3x-|x+15|=5/4 `

` =>|x+15|=3x-5/4 `

` ĐKXĐ:3x-5/4≥0=>3x≥5/4=>x≥5/12 `

` => ` \(\left[ \begin{array}{l}x+15=3x-\dfrac{5}{4}\\x+15=-3x+\dfrac{5}{4}\end{array} \right.\)

` => ` \(\left[ \begin{array}{l}x-3x=-\dfrac{5}{4}-15\\x+3x=\dfrac{5}{4}-15\end{array} \right.\)

` => `  \(\left[ \begin{array}{l}-2x=-\dfrac{65}{4}\\4x=-\dfrac{55}{4}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x=\dfrac{65}{8}(TMĐK)\\x=\dfrac{-55}{16}(KTMĐK)\end{array} \right.\) 

` d)3/2-|5/4+3x|=1/4 `

` =>|5/4+3x|=3/2-1/4=5/4 `

` => ` \(\left[ \begin{array}{l}\dfrac{5}{4}+3x=-\dfrac{5}{4}\\\dfrac{5}{4}+3x=\dfrac{5}{4}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}3x=-\dfrac{5}{2}\\3x=0\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x=-\dfrac{5}{6}\\x=0\end{array} \right.\) 

` e)|4x-1|=|3x-1/2| `

` => ` \(\left[ \begin{array}{l}4x-1=3x-\dfrac{1}{2}\\4x-1=-3x+\dfrac{1}{2}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}4x-3x=-\dfrac{1}{2}+1\\4x+3x=\dfrac{1}{2}+1\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{array} \right.\) ` => ` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{array} \right.\) 

` f)|2x-1|=|x+1/3| `

` => ` \(\left[ \begin{array}{l}2x-1=x+\dfrac{1}{3}\\2x-1=-x-\dfrac{1}{3}\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}2x-x=\dfrac{1}{3}+1\\2x+x=-\dfrac{1}{3}+1\end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}x=\dfrac{4}{3}\\3x=\dfrac{2}{3}\end{array} \right.\) ` => ` \(\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=\dfrac{2}{9}\end{array} \right.\)