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\(\left(2x-6\right)^5=\left(2x-6\right)^2\)
\(\Rightarrow\left(2x-6\right)^5-\left(2x-6\right)^2=0\)
\(\Rightarrow\left(2x-6\right)^2\left[\left(2x-6\right)^3-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=6\\2x-6=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{7}{2}\end{cases}}\)

a)\(5^x=125\)
\(\Rightarrow5^x=5^3\Rightarrow x=3\)
b)\(3^{2x}=81\)
\(\Rightarrow3^{2x}=3^4\)
\(\Rightarrow2x=4\Rightarrow x=2\)
c)\(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
\(\Rightarrow5^{2x-3}-2\cdot25=25\cdot3\)
\(\Rightarrow5^{2x-3}-50=75\)
\(\Rightarrow5^{2x-3}=125\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\Rightarrow x=3\)

a) Ta có : 5x = 125
=> 5x = 53
=> x = 3
b) Ta có : 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 4 : 2 = 2
c) Ta có : 52x + 3 - 2 . 52 = 52 . 3
=> 52x + 3 = 52 . 3 + 2 . 52
=> 52x + 3 = 52 . (3 + 2)
=> 52x + 3 = 52 . 5 = 53
=> 2x + 3 = 3
=> 2x = 3 - 3 = 0
=> x = 0 : 2 = 0
a)5x=125<=>5x=53=>x=3
b)32x=81<=>32x=34=>2x=4=>x=2
c)52x+3-50=75
52x+3=75+50
52x+3=125
<=>52x+3=53
=>2x+3=3
=>2x=0
=>x=0


(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

(=) 52x-3 -2 =3
(=) 52x-3 = 5
(=)2x-3=1
(=)2x=4
(=)x=2
#Học-tốt