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1 tháng 9 2015

(2x-15)5=(2x-15)3

=>(2x-15)5-(2x-15)3=0

=>(2x-15)3.(2x-15)2-(2x-15)3.1=0

=>(2x-15)3.((2x-15)2-1)=0

=>(2x-15)3=0=>2x-15=0=>2x=15=>x=15/2

hoặc (2x-15)2-1=0=>(2x-15)2=0=>2x-15=1,-1=>2x=16,14=>x=8,7

Vậy x=15/2,8,7.

1 tháng 9 2015

ta co ( 2x-15)5= (2x-15)3

=> (2x-15)5-(2x-15)3=0

=> (2x-15)3 .{(2x-15)2-1}=0

=> (2x-15)3=0 hoac (2x-15)2-1=0

=> 2x=15 hoac (2x-15)2=1

=> x=15/2 hoac 2x-15=-1;1

=> x=15/2 hoac 2x= 14;16 => x=7;8

vay x=15/2;7;8

 

10 tháng 2 2021

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\) 

Vậy.........

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

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

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

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)

\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)

27 tháng 12 2022

Tham khảo : 

`(2x - 15)^5 = (2x - 15)^3`

`=> (2x - 15)^5 : (2x - 15)^3 = 1`

`=> (2x - 15)^2 = 1`

`=> (2x - 15)^2 = 1^2`

`=>` $\left[\begin{matrix} 2x-15=1\\ 2x-15=-1\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x=1 + 15\\ 2x=-1 + 15\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2x=16\\ 2x=14\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=8\\ x=7\end{matrix}\right.$

`=> x in {7;8}`

9 tháng 4 2023

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

\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)

\(=>x\in\left\{\dfrac{15}{2};8\right\}\)

1 tháng 1 2021

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

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

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

\(\Leftrightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

\(\Leftrightarrow x=\frac{15}{2};8;7\)

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