a)x+(-19)=27 b)x+34=-18 c)-129-x=42 d)37-/x/=16 e)-x+23=-19 f)42-/x+2/=23

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1. a, $x+(-19)=27$

   $x=27+19$

   $x=46$

   b, $x+34=-18$

   $x=-18+(-34)$

   $x=-52$

   c,$-129-x=42$

   $x=-129+(-42)$

   $x=-171$

   d, $37-|x|=16$

   $→x=21$

   $→x∈{21,-21}$

   Vậy $x∈{21,-21}$

   e, $-x+23=-19$

   $-x=-19+(-23)$

   $-x=-42$

   $→x=42$

   f, $42-|x+2|=23$

   $→ |x+2|=19$

   $→ (x+2)∈{19,-19}$

   TH1: $x+2=19$

   $x=17$

   TH2: $x+2=-19$

   $x=-21$

   Vậy $x∈{17,-21}$

    

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2. Đáp án:

   `a)x=46`

   `b)x=-52`

   `c)x=-171`

   `d)x∈` { `21;-21` }

   `e)x=42`

   `f)x∈` { `17;-21` }

   Giải thích các bước giải:

   Ta có :

   `a,x+(-19)=27`

   `→x=27-(-19)`

   `→x=46`

   `———–`

   `b,x+34=-18`

   `→x=-18-34`

   `→x=-52`

   `———–`

   `c,-129-x=42`

   `→x=-129-42`

   `→x=-171`

   `————`

   `d,37-|x|=16`

   `→|x|=37-16`

   `→|x|=21`

   `→` \(\left[ \begin{array}{l}x=21\\x=-21\end{array} \right.\) 

   `→x∈` { `21;-21` }

   `————`

   `e,-x+23=-19`

   `→-x=-19-23`

   `→-x=-42`

   `→x=42`

   `f,42-|x+2|=23`

   `→|x+2|=42-23`

   `→|x+2|=19`

   `→` \(\left[ \begin{array}{l}x+2=19\\x+2=-19\end{array} \right.\) 

   `→` \(\left[ \begin{array}{l}x=17\\x=-21\end{array} \right.\) 

   `→x∈` { `17;-21` }

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