(2x-1)mũ 6 = (2x-1)mũ5
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H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
\(27^5:9^6\)
\(=\left(3^3\right)^5:\left(3^2\right)^6\)
\(=3^{15}:3^{12}\)
\(=3^{15-12}\)
\(=3^3\)
24 tháng 9 2021
\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\\ \Rightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
24 tháng 9 2021
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=-1\\2x-1=1\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
16 tháng 1 2022
\(\left(2x+1\right)^4=\left(2x+1\right)^6\\ \Rightarrow\left(2x+1\right)^6-\left(2x+1\right)^4=0\\ \Rightarrow\left(2x+1\right)^4\left[\left(2x+1\right)^2-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\2x+1=1\\2x+1=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)
DA
1
16 tháng 9 2016
\(\left(\frac{1}{5}\right)^5\cdot5^5\)
\(=\left(\frac{1}{5}\cdot5\right)^5\)
\(=1^5\)
\(=1\)
10 tháng 9 2020
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
XO
6 tháng 10 2020
(2x - 1)6 = (2x - 1)4
=> (2x - 1)6 - (2x - 1)4 = 0
=> (2x - 1)4.[(2x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=\pm1\end{cases}}\)
Khi 2x - 1 = 0 => x = 1/2
Khi 2x - 1 = -1 => x = 0
Khi 2x - 1 = 1 => x = 1
Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)là giá trị cần tìm
6 tháng 10 2020
( 2x - 1 )6 = ( 2x - 1 )4
<=> ( 2x - 1 )6 - ( 2x - 1 )4 = 0
<=> ( 2x - 1 )4[ ( 2x - 1 )2 - 1 ] = 0
<=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)( thay = dấu hoặc hộ nhé )
6 tháng 8 2017
1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
\(\left(2x-1\right)^6=\left(2x-1\right)^5\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^5=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left[\left(2x-1\right)-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(2x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\).