Toán Lớp 7: | 3x – 1/9 | – 2/81 = [(1/3)^2]^3 : ( 1/3)^2

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1. Giải đáp:

   | 3x – 1/9 | – 2/81 = [(1/3)^2]^3 : ( 1/3)^2

   | 3x – 1/9 | – 2/81 = (1/3)^6 : (1/3)^2

   | 3x – 1/9 | – 2/81 = (1/3)^4

   | 3x – 1/9 | – 2/81 = 1/81

   | 3x – 1/9 | = 1/81 + 2/81 = 3/81

   | 3x – 1/9 | = 1/27

   -> Ta chia 2 trường hợp :

   Th1 : 3x – 1/9 = 1/27

   -> 3x = 4/27

   -> x = 4/81

   Th2 : 3x – 1/9 = -1/27

   3x = 2/27

   -> x = 2/81

   Vậy x in { 4/81 ; 2/81 }

   Trả lời
2. Giải đáp và Lời giải và giải thích chi tiết:

   ${| 3x – \dfrac1 9 |-}$ $\dfrac{2}{81}=$ $ [(\dfrac{1}{3})^2]^3:$ $(\dfrac{1}{3})^2$

   $=|3x-\dfrac{1}{9}|-$ $\dfrac{2}{81}=$ $\dfrac{(\frac{1}{3})^6}{(\frac{1}{3})^2}$

   $=|3x-\dfrac{1}{9}|-$ $\dfrac{2}{81}=$ $(\dfrac{1}{3})^4$

   $=|3x-\dfrac{1}{9}|-$ $\dfrac{2}{81}$ $=\dfrac{1}{81}$

   $=|3x-\dfrac{1}{9}|-$ $\dfrac{2}{81}-$ $\dfrac{1}{81}=0$

   $=|3x-\dfrac{1}{9}|-$ $\dfrac{1}{27}=0$

   $=|3x-\dfrac{1}{9}|=$ $\dfrac{1}{27}$

   $3x-\dfrac{1}{9}=$ $\dfrac{1}{27}$

   $3x-\dfrac{1}{9}=$ $-\dfrac{1}{27}$

   $3x=\dfrac{4}{27}$

   $3x=\dfrac{2}{27}$

   $x=\dfrac{4}{81}$

   $x=\dfrac{2}{81}$ 

   #ThiênShi

   Trả lời