Tìm x biết a,|x+5|=100 b,100 – | 25-x | = 40 c,|2x+1| = | 23-x | d,|11-7x| – | 4x + 3|=0

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1. a. $|x+5|=100$

   ⇒ \(\left[ \begin{array}{l}x+5=100\\x+5=-100\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=100-5\\x=-100-5\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=95\\x=-105\end{array} \right.\) 

   b. $100-| 25-x | = 40$

   ⇒ $| 25-x | =100-40$

   ⇒ $| 25-x | =60$

   ⇒ \(\left[ \begin{array}{l}25-x=60\\25-x=-60\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=25-60\\x=25-(-60)\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=-35\\x=85\end{array} \right.\) 

   c. $|2x+1| = | 23-x |$

   ⇒ \(\left[ \begin{array}{l}2x+1=23-x\\2x+1=-(23-x)\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}2x+x=23-1\\2x+1=-23+x\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}3x=22\\2x-x=-23-1\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}x=\frac{22}{3}\\x=-24\end{array} \right.\) 

   d. $|11-7x| – | 4x + 3|=0$

   ⇒ $|11-7x| = | 4x + 3|$

   ⇒ \(\left[ \begin{array}{l}11-7x=4x+3\\11-7x=-(4x+3)\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}-7x-4x=3-11\\11-7x=-4x-3\end{array} \right.\) 

   ⇒ \(\left[ \begin{array}{l}-11x=-8\\4x-7x=-3-11\end{array} \right.\) 

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

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

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2. Giải thích các bước giải:

   a,|x + 5| = 100    

   => x + 5 = -100

         x + 5 = 100

   TH1: x + 5 = -100    

             x        = (-100) – 5

             x        = -105

   TH2 :  x + 5 = 100

              x         = 100 – 5

              x          = 95

   b,100 – I25 – xI  = 40

   100 – 40              = I25 – xI

        60                    = I25 – xI

   => 25 – x = 60

         25 – x = -60

   TH1: 25 – x = 60

                    x = 25 – 60

                    x = -35

   TH2 : 25 – x = -60

                      x = 25 – (-60)

                      x = -85

    

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