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

0 bình luận về “|x-1|=|2|+|-6| 3/4+ |x-5/2| = 1/4 3/2-|x-2/3| = 4/3 |x-2|= |-1/2| + |-3/4|”

1.  | x-1 | = |2| + | -6 |

    | x-1 | = 2+ 6

    | x-1 | = 8

   => x-1 = 8 hoặc x -1 = -8

         x = 8 + 1        x = (-8 ) + 1

         x = 9                x = ( -7 )

   vậy x ∈ { 9 ; -7 }

   3/4 + | x – 5/2 | = 1/4 

   | x – 5/2 | = 1/4 – 3/4 

   | x – 5/2 | = -1/2

   vì | x – 5/2 | ≥  0 mà -1/2 < 0

   => x ∈ ∅

   Vậy x ∈ ∅

   3/2 – | x -2/3 | = 4/3 

   | x – 2/3 | = 3/2 – 4/3

   | x – 2/3 | = 1/6

   => x – 2/3 = 1/6 hoặc x – 2/3 = -1/6

         x = 1/6 + 2/3          x = -1/6 + 2/3

         x = 5/6                     x = 1/2

   vậy x ∈ { 5/6 ; 1/2}

   | x – 2 | = | -1/2 | + | -3/4 |

   | x – 2 | = 1/2 + 3/4

   | x- 2 | = 5/4

   => x – 2 = 5/4 hoặc x – 2 = -5/4

         x =  5/4 + 2        x = -5/4 + 2 

         x = 13/4              x = 3/4

   vậy x ∈ { 3/4 ; 13/4 }

    

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2. Đáp án:

    

   Giải thích các bước giải:

    `|x- 1|= |2|+ |-6|`

   `|x+ 1|= -2+ 6`

   `|x+ 1|= 4`

   `⇒ |x+ 1|= ±4`

   `Th1: x+ 1= -4`

   `x= -4- 1`

   `x= -5`

   `Th2: x+ 1= 4`

   `x= 4- 1`

   `x= 3`

   Vậy `x∈ \text{{-5; 3}}`

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

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

   `|x- 5/2|= -1/2`

   `x- 5/2= -1/2`

   `x=  -1/2+ 5/2`

   `x= 4/2= 2`

   `3/2- |x- 2/3|= 4/3`

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

   `|x- 2/3|= 8/6+ 9/6`

   `|x- 2/3|= 17/6`

   `|x+ 2/3|= ±17/6`

   `Th1: x+ 2/3= 17/6`

   `x= 17/6- 2/3`

   `x= 17/6- 4/6`

   `x= 13/6`

   `Th2: x+ 2/3= -17/6`

   `x= -17/6- 2/3`

   `x= -17/6+ 4/6`

   `x= -13/6`

   Vậy `x∈ \text{{-13/6; 13/6}}`

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

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

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

   `|x- 2|= 5/4`

   `⇒ |x- 2|= ±5/4`

   `Th1: x- 2= 5/4`

   `x= 5/4+ 2`

   `x= 5/4+ 8/4`

   `x= 13/4`

   `Th2: x- 2= -5/4`

   `x= -5/4+ 2`

   `x= -5/4+ 8/4`

   `x= 3/4`

   Vậy `x∈ \text{{3/4; 13/4}}`

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