-1`1/2`-x=`-2/3`+0,75 I`4/5`-xI-`1/10`=`2/5`

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1. `-1 1/2 – x = ( – 2 )/3 + 0,75`

   `⇔ ( – 3 )/2 – x = ( – 2 )/3 + 3/4`

   `⇔ ( – 3 )/2 – x =  1/12`

   `⇔                 x = ( – 3 )/2 – 1/12`

   `⇔                 x = ( – 19 )/12`

   Vậy , `x = ( – 19 )/12`

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   `| 4/5 – x | – 1/10 = 2/5`

   `⇔   | 4/5 – x |     = 2/5 + 1/10`

   `⇔   | 4/5 – x |     = 1/2`

   Trường hợp `1 : 4/5 – x = 1/2`

   `⇔ x = 4/5 – 1/2`

   `⇔ x = 3/10`

   Trường hợp `2 :  4/5 – x = ( – 1 )/2`

   `⇔ x = 4/5 – ( – 1 )/2`

   `⇔ x =  13/10`

   Vậy , `x ∈ { 3/10 ; 13/10 }`

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2. Giải thích các bước giải:

   `-1 1/2-x=-2/3+0,75`
   `=>-3/2 -x=-2/3+3/4`
   `=>-3/2-x=-8/12+9/12`
   `=>-3/2-x=1/12`
   `=>x=-3/2-1/12`
   `=>x=-18/12-1/12`
   `=>x=-19/12`
   Vậy `x=-19/12`
   `|4/5-x|-1/10=2/5`
   `=>|4/5-x|=2/5+1/10`
   `=>|4/5-x|=4/10+1/10`
   `=>|4/5-x|=5/10`
   `=>|4/5-x|=1/2`
   TH`1`:
   `4/5-x=1/2`
   `=>x=4/5-1/2`
   `=>x=3/10`
   TH`2`
   `4/5-x=-1/2`
   `=>x=4/5+1/2`
   `=>x=13/10`
   Vậy `\in{3/10;13/10}`

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