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) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

a,  2 x - 15 = 17

⇒ 2 x = 2 = 2 2 ⇒ x = 1

b,  7 x - 11 3 = 2 5 . 5 2 + 200

⇒ 7 x - 11 3 = 1000 = 10 3

⇒ 7x – 11 =10

⇒ 7x = 21

⇒ x = 3

Ta có : \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\2x=16;14\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=8;7\end{cases}}\)

\(\Rightarrow\left(2x-15\right)^5=\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7,5\\x=8;7\end{cases}}\)

25/7nha

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 

tìm số tự nhiên x biết

a, 2^x - 15 =17

=> 2^x = 17 + 15 = 32

=> 2^x = 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

b: =>15-x=-10

hay x=25

a: =>-2x+17=9

=>-2x=-8

hay x=4

d: \(\Leftrightarrow9x^2=81\)

hay \(x\in\left\{3;-3\right\}\)

e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)