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

By Uyên Thi

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

### 0 bình luận về “Toán Lớp 7: | 3x – 1/9 | – 2/81 = [(1/3)^2]^3 : ( 1/3)^2”

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