a ) 2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
Vậy x = 5
b ) x15 = x
=> x15 : x = 1
x14 = 1
x14 = 114
=> x = 1
Vậy x = 1
c ) ( 2x + 1 )3 = 125
( 2x + 1 )3 = 53
=> 2x + 1 = 5
2x = 5 - 1
2x = 4
=> x = 4 : 2
x = 2
Vậy x = 2
d ) ( x - 5 )4 = ( x - 5 )6
( x - 5 )6 - ( x - 5 )4 = 0
( x - 5 )4 . [ ( x - 5 )2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\text{[}\left(x-5\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x-5=1\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
Vậy x thuộc { 5 ; 6 }
a/ 2^x .4 =128
2^x =128 : 4
2^x = 32
mak 2^4 =32 => x= 5
b/ x^15 = x
mak ta có 0 mũ bao nhiêu cx = 0 => x=0
c/ ( 2x + 1 )^3 = 125
mak 5^3 =125
ta có 2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2 = 2
Làm nốt câu c =.=
(x-5)^4 = (x-5)^6
(x-5)^4 = [(x-5)^4]*[(x-5)^2]
[(x-5)^4] : [(x-5)^4] = (x-5)^2
(x-5)^2 = 1
TH1 (x-5)^2 = (-1)^2
=> (x-5) = -1 => x= -1 + 5 => x=4
TH2 (x-5)^2 = 1^2
=> (x-5) = 1 => x = 1 + 5 => x=6