2.tìm x,biết: a,3.x-(2.x-4(-5))=4 b,8./x-1/-5=11 c,5-2 /2.x+1/=1

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1. a, 3.x –  [2.x – 4 .( – 5)] = 4

   <=> 3.x – (2.x + 20) = 4 

   <=> 3.x – 2.x – 20 = 4

   <=> x = 24 

   b, 8.|x – 1| – 5 = 11 

   <=> 8.|x – 1| = 16 

   <=> |x – 1| = 2 

   <=> x – 1 = 2 hoặc x – 1 = – 2

   <=> x = 3 hoặc x = – 1

   c, 5 – 2.|2.x + 1| = 1

   <=> 2.|2.x + 1| = 4 

   <=> |2.x + 1| = 2 

   <=> 2.x + 1 = 2 hoặc 2.x + 1 = – 2

   <=> 2.x = 1 hoặc 2.x = – 3

   <=> x = 1/2 hoặc x = – 3/2 

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2. `a) 3x-(2x-4.(-5))=4`

   `<=> 3x-(2x+20)=4`

   `<=> 3x-2x-20=4`

   `<=> x=4+20`

   `<=> x=24`

   Vậy `x=24`

   `b) 8|x-1|-5=11`

   `<=>8|x-1|=16`

   `<=> |x-1|=2`

   `<=>`\(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\) 

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

   Vậy `x∈{3;-1}`

   `c) 5-2|2x+1|=1`

   `<=> 2|2x+1|=4`

   `<=> |2x+1|=2`

   `<=>`\(\left[ \begin{array}{l}2x+1=2\\2x+1=-2\end{array} \right.\) 

   `<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{-3}{2}\end{array} \right.\) 

   Vậy `x∈{1/2;-3/2}`

    

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