|x+13|-|-10|=-8+2 |x-11|-(-5+6)=-|-5| |x+11|-|-3|+|-5|=8 |x-3|-(5-7+|-3|=|5-7| 5-|x-1|=3-|4+5-7| 51-|x-1|=-(4-51+3| -12+3-|x+2|=7-12 14-(5-6)-|x-3|=14

Question

|x+13|-|-10|=-8+2
|x-11|-(-5+6)=-|-5|
|x+11|-|-3|+|-5|=8
|x-3|-(5-7+|-3|=|5-7|
5-|x-1|=3-|4+5-7|
51-|x-1|=-(4-51+3|
-12+3-|x+2|=7-12
14-(5-6)-|x-3|=14
23-(7-4)-|x+7|=1-(-23)
Thanks

in progress 0
Adalyn 1 tháng 2021-11-08T21:37:07+00:00 1 Answers 4 views 0

Answers ( )

    0
    2021-11-08T21:38:23+00:00

    `a)|x+13|-|-10|=-8+2`

    `→|x+13|-10=-6`

    `→|x+13|=-6+10`

    `→|x+13|=4`

    `→` \(\left[ \begin{array}{l}x+13=4\\x+13=-4\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=-9\\x=-17\end{array} \right.\) 

    Vậy `x∈{-9;-17}`

    `b)|x-11|-(-5+6)=-|-5|`

    `→|x-11|-1=-5`

    `→|x-11|=-4`

    `|x-11|≥0→x∈∅`

    Vậy `x∈∅`

    `c)|x+11|-|-3|+|-5|=8`

    `→|x+11|-3+5=8`

    `→|x+11|=8+3-5`

    `→|x+11|=6`

    `→` \(\left[ \begin{array}{l}x+11=6\\x+11=-6\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=-5\\x=-17\end{array} \right.\) 

    Vậy `x∈{-5;-17}`

    `d)|x-3|-(5-7+|-3|=|5-7|`

    `→|x-3|-1=2`

    `→|x-3|=3`

    `→` \(\left[ \begin{array}{l}x-3=3\\x-3=-3\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=6\\x=0\end{array} \right.\) 

    Vậy `x∈{6;0}`

    `e)5-|x-1|=3-|4+5-7|`

    `→5-|x-1|=1`

    `→|x-1|=4`

    `→` \(\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\) 

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

    Vậy `x∈{5;-3}`

    `f)51-|x-1|=-(4-51+3)`

    `→51-|x-1|=44`

    `→|x-1|=7`

    `→` \(\left[ \begin{array}{l}x-1=7\\x-1=-7\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=8\\x=-6\end{array} \right.\) 

    Vậy `x∈{8;-6}`

    `g)-12+3-|x+2|=7-12`

    `→-9-|x+2|=-5`

    `→|x+1|=-9-(-5)`

    `→|x+1|=-4`

    `|x+1|≥0→x∈∅`

    `h)14-(5-6)-|x-3|=14`

    `→15-|x-3|=14`

    `→|x-3|=15-14`

    `→|x-3|=1`

    `→` \(\left[ \begin{array}{l}x-3=1\\x-3=-1\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) 

    Vậy `x∈{4;2}`

    `i)23-(7-4)-|x+7|=1-(-23)`

    `→20-|x+7|=24`

    `→|x+7|=20-24`

    `→|x+7|=-4`

    `|x+7|≥0→x∈∅`

     

Leave an answer

Browse

35:5x4+1-9:3 = ? ( )