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![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( -2x - 28 )5 = -32 = ( -2 )5
-2x - 28 = -2
-2x = ( -2 ) + 28 = 26
x = 26 : ( -2 ) = -13
b) ( 7x - 11 )3 = 1000 = 103
7x - 11 = 10
7x = 10 + 11 = 21
x = 21 : 7 = 3
c) ( 3x - 15 )2 - 6 = 75
( 3x - 15 )2 = 75 + 6 = 81 = 92
3x - 15 = 9
3x = 9 + 15 = 24
x = 24 : 3 = 8
![](https://rs.olm.vn/images/avt/0.png?1311)
tìm số tự nhiên x biết
a, 2x - 15 =17
=> 2x = 17 + 15 = 32
=> 2x = 25
=> x = 5
b, ( 7x - 11)3 = 25 x 52+200
=> ( 7x - 11 )3 = 32 . 25 + 200
=> ( 7x - 11 )3 = 1000
=> ( 7x - 11 )3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11 = 21
=> x = 21 : 7 = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
(7x-11)3=25.52+200
=> (7x-11)3=800+200
=> (7x-11)3=1000
=> (7x-11)3=103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
(2x-15)5=(2x-15)3
=> (2x-15)5 - (2x-15)3 = 0
=> (2x-15)3 . [ (2x-15)2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}2x=15\\2x=16\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
Mà x thuộc N
=> x = 8
(3x-5)10=(3x-5)9
=> (3x-5)10 - (3x-5)9 = 0
=> (3x-5)9 .[ (3x-5) - 1 ] = 0
=> \(\orbr{\begin{cases}\left(3x-5\right)^9=0\\\left(3x-5\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-5=0\\3x-5=1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=5\\3x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
Mà x thuộc N
=> x = 2
a)(7x-11)^3=1000
(7x-11)^3=10^3
7x-11 =10
7x =10+11=21
x =21:7=3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 2x - 15 = 17
2x =17+15
2x =32
x=8
b,{7x . 11}3 = 25 .52 +200
=>{7x . 11}3= 32.25+200
=>{7x . 11}3= 800+200
=>{7x . 11}3=1000
VÌ 103 =1000=> 7x.11=10
\(7x=\frac{10}{11}\)
=> x=\(\frac{10}{77}\)
c, x10 =1x
=> x=1
d,x10=1
=>x=1
e,2x-15=1
=>2x=1+15
2x=16
=>x=8
a. 2x - 15 = 17
2x = 17 + 15
2x = 32
=> x = 5
Vậy : 25 - 15 = 17
![](https://rs.olm.vn/images/avt/0.png?1311)
b) x^10=1x
=> x=0 ; 1
c)(2x-15)^5=(2x-15)^3
=> (2x-15)^4 = (2x-15)^2
=> (2x-15)^2 =1
=> 2x-15=1
=> x=8
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)