Tìm x t Z biết a) -12(x-5)+7(3-x)=5 b)30(x+2)-6(x-5)-27x=100

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1. a) -12(x-5)+7(3-x)=5

   ⇔ -12x +60 +21 -7x = 5

   ⇔ -12x -7x = 5 -60 -21

   ⇔ -19x = -76

   ⇔ x = 4

     Vậy x= 4

   b)30(x+2)-6(x-5)-27x=100

   ⇔ 30x +60 -6x +30 -27x = 100

   ⇔ 30x -6x -27x = 100 -60 -30

   ⇔ -3x = 10

   ⇔ x = -10/3

     Vậy x= -10/3

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2. a) $-12(x-5) + 7(3-x) = 5$
   $⇔ -12x + 60 + 21 – 7x = 5$ 
   $⇔ -12x – 7x = 5 – 60 – 21$
   $⇔ -19x = -76$
   $⇔ x = ( – 76 ) ÷ ( -19 )$
   $⇔ x = 4$

   b) $30(x+2) -6(x-5)-27x=100$
   $⇔ 30x + 60 -6x + 30 – 27x = 100$
   $⇔ 30x – 27x – 6x = 100 – 30 – 60$
   $⇔ -3x = 10$
   $⇔ x = \dfrac{-10}{3}$

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