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a) xem lại đề
b) 3x-1=27
=>3x-1=33
=>x-1=3
=>x=3+1
=>x=4
c)3x+1=9
=>3x+1=32
=>x+1=2
=>x=2-1
=>x=1
a) x2=x3
⇒ x2-x3=0
⇒ x2-x2.x=0
⇒ x2.(x+1)=0
⇒\(\left\{{}\begin{matrix}\text{x}^2=0\\\text{x}+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\text{x}=0\\\text{x}=0+1=1\end{matrix}\right.\)
⇒ \(\text{x}\in\left\{0;1\right\}\)
b) 3x-1=27
⇒ 3x-1=33
⇒ x-1=3
⇒ x= 3+1=4
c) 3x+1=9
⇒ 3x+1= 33
⇒ x+1=3
⇒ x=3-1=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (3x-1)3 = 27 = 33
=> 3x - 1 = 3
3x = 4
x = 4/3
b) 32x-1 = 27 = 33
=> 2x - 1 = 3
2x = 4
x = 2
c) 68-2x = 36 = 62
=> 8 -2x = 2
=> 2x = 6
x = 3
d) (3x-2)10 = (3x-2)7
=> (3x-2)10 - (3x-2)7 = 0
(3x-2)7. [(3x-2)3 - 1] = 0
=> (3x-2)7 = 0 => 3x-2 = 0 => 3x = 2 => x = 2/3
(3x-2)3 - 1 = 0 => (3x-2)3 = 1 => 3x - 2 = 1 => 3x = 3 => x = 1
KL:...
![](https://rs.olm.vn/images/avt/0.png?1311)
-1 5/27-(3x-7/9)3=-24/27
(3x-7/9)3 =-32/27-(-24/27)
(3x-7/9)3=-8/27
(3x-7/9)3=(-2/3)3
⇒3x-7/9=-2/3
3x =-2/3+7/9
3x =1/9
x =1/9:3
x =1/27
Chúc bạn học tốt!
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+
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33543537656453664
![](https://rs.olm.vn/images/avt/0.png?1311)
-1 5/27-(3x-7/9)3=-24/27
(3x-7/9)3 =-32/27-(-24/27)
(3x-7/9)3=-8/27
(3x-7/9)3=(-2/3)3
⇒3x-7/9=-2/3
3x =-2/3+7/9
3x =1/9
x =1/9:3
x =1/27
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow3x-1=-\dfrac{2}{3}\Rightarrow3x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-1 . \(\dfrac{5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = -\(\dfrac{24}{27}\)
\(\Rightarrow\) \(\dfrac{-5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-24}{27}\)
\(\Rightarrow\) \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-5}{27}\) - \(\dfrac{-24}{27}\) = \(\dfrac{19}{27}\)
\(\Rightarrow\) \(3^3\) . \(\dfrac{7^3}{3^6}\) = \(\dfrac{19}{3^3}\)
\(\Rightarrow\) \(\dfrac{x^3.7^3}{3^3}\) = \(\dfrac{19}{3^3}\)
\(\Rightarrow\) \(x^3\) . \(7^3\) = 19
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: 3 x = 3 2 nên x = 2
b, Ta có: 5 x = 5 3 nên x = 3
c, Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1
d, Ta có: 6 x - 1 = 6 2 nên x - 1 = 2, đo đó x = 3
e) Ta có: 3 2 x + 1 = 3 3 nên 2x +1 = 3, do đó x = 1
f) Ta có: x 50 = x nên x 50 - x = 0 , do đó x x 49 - 1 = 0 = 0
Vì thế x = 0 hoặc x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
( 3x - 1 )3 = 27
( 3x - 1 )3 = 33
3x - 1 = 3
3x = 3 + 1
3x = 4
x = 4 : 3
3x-1