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a) (2x-1)5 = 2x-1
=> (2x-1)5 - (2x-1) = 0
(2x-1).[(2x-1)4 -1]=0
=> 2x-1 =0 => 2x = 1=> x = 1/2
(2x-1)4 - 1= 0 => (2x-1)4 = 1 = 14 = (-1)4
=> 2x-1= 1 => 2x = 2 => x = 1
2x -1 = -1 => 2x = 0 => x = 0
KL:...
b) 25x.43 = 169
25x = (42)9:43
25x = 418:43
25x = 415 = (22)15 = 230
=> 5x = 30 => x = 6
c) 3x+1+4.3x = 7.36
3x.3+4.3x = 5013
3x.(3+4) = 5013
3x.7 = 5013
3x = 729 = 36
=> x = 6

(2x - 6)5 = (2x - 6)2
=> (2x - 6)5 - (2x - 6)2 = 0
=> (2x - 6)2.[(2x - 6)3 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=6\\2x-6=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\2x=7\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{7}{2}\left(ktm\right)\end{cases}}\)
33x - 4 - x0 = 8
=> 33x - 4 - 1 = 8
=> 33x - 4 = 8 +1
=> 33x - 4 = 9
=> 33x - 4= 32
=> 3x - 4 = 2
=> 3x = 2 + 4
=> 3x = 6
=> x = 6 : 3 = 2

\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)

a, 2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b, x15 = x1
=> x15 - x = 0
x . ( x14 - 1 ) = 0
=> x = 0 hoặc x14 - 1 = 0
=> x = 0 hoặc x = 1
c, (2x + 1)3 = 125
( 2x + 1 )3 = 53
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 4 : 2
=> x = 2
d, (x – 5)4 = (x - 5)6
=> ( x - 5 )6 - ( x - 5 )4 = 0
=> ( x - 5 )4 . [ ( x - 5 )2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
e, x10 = x
x10 - x = 0
x . ( x9 - 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
f, (2x -15)5 = (2x -15)3
( 2x - 15 )5 - ( 2x - 15 )3 = 0
( 2x - 15 )3 . [ ( 2x - 15 )2 - 1 ] = 0
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}x\text{ không tồn tại}\\x=8\end{cases}}}\)
Tìm x , biết:
a, ( 2x + 1)2 = 93
b, ( 2x + 1)3 = 1252
c, 252< 5x+3 < 254
d, x15= x
e, ( x - 5)4 = ( x-5)6

A) \(\left(2x+1\right)^2=9^3\)
\(\Rightarrow\left(2x+1\right)^2=9^2\times9\)
\(\Rightarrow2x+1=81\)
\(\Rightarrow2x=81-1\)
\(\Rightarrow2x=80\)
\(\Leftrightarrow x=40\)
B) \(\left(2x+1\right)^3=125^2\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Leftrightarrow x=2\)
C) \(25^2< 5^{x+3}< 25^4\)
\(\Leftrightarrow5^4< 5^{x+3}< 5^6\)
\(\Leftrightarrow4< x+3< 6\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
D) \(x^{15}=x\)
Nếu \(x>1\)thì \(x^{15}>x\)
Vậy \(x=1\)
E) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)\times\left(1^4-1^6\right)=0\)
\(\Rightarrow x-5=0\)
\(\Leftrightarrow x=5\)
KÍCH MK NHA BẠN
\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
\(\Rightarrow\left(2x-3\right)\in\left\{1;-1;0\right\}\)
( 2x - 3 )4 = ( 2x - 3 )6
TH1:
2x - 3 = 0
2x = 3
x = 3/2
TH2:
2x - 3 = 1
2x = 1 + 3
2x = 4
x = 4 : 2
x = 2
TH3:
2x - 3 = -1
2x = -1 + 3
2x = 2
x = 2 : 2
x = 1
Vậy x thuộc {3/2 ; 1 ; 2}
Chúc bạn học tốt ^^