1/ Tìm x : a) |x + 3| = 11 b) |x – 4| = 5 + 8

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1. `1)`

   `a) |x + 3| = 11`

   `=> x + 3 = 11 hoặc -11`

   `TH1 : x + 3 = 11`

   `=> x = 11 – 3`

   `=> x = 8.`

   `TH2 : x + 3 = -11`

   `=> x = -11 – 3`

   `=> x = -14.`

   `b) |x – 4| = 5 + 8`

   `|x – 4| = 13`

   `=> x – 4 = 13 hoặc -13`

   `TH1 : x – 4 = 13`

   `=> x = 13 + 4`

   `=> x = 17`

   `TH2 : x – 4 = -13`

   `=> x = -13 + 4`

   `=> x = -9.`

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2. 1/ 

   a) |x + 3| = 11

   ⇒\(\left[ \begin{array}{l}x+3=11\\x+3=-11\end{array} \right.\) 

   ⇒\(\left[ \begin{array}{l}x=11-3=8∈Z\\x=-11-3=-14∈Z\end{array} \right.\)

   Vậy x∈{8; -14}

   b) |x – 4| = 5 + 8

   ⇒|x – 4| =13

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

   ⇒\(\left[ \begin{array}{l}x=13+4=17∈Z\\x=-13+4=-9∈Z\end{array} \right.\) 

   Vậy x∈{17; -9}

    

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