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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 : \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\2x=16;14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=8;7\end{cases}}\)
\(\Rightarrow\left(2x-15\right)^5=\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7,5\\x=8;7\end{cases}}\)
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, 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
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: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
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