\(\left(2x-1\right)^8=\left(2x-1\right)^6\)
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NH
1
FT
12 tháng 10 2017
\(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(A=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(A=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(A=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(A=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(A=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(A=\dfrac{5^{32}-1}{2}\)
FT
12 tháng 10 2017
\(A=\left|2x+4\right|+\left|2x+6\right|+\left|2x+8\right|\)
\(A=\left|2x+4\right|+\left|2x+8\right|+\left|2x+6\right|\)
\(A=\left|2x+4\right|+\left|-2x-8\right|+\left|2x+6\right|\)
\(A\ge\left|2x+4-2x-8\right|+\left|2x+6\right|\)
\(A\ge4+\left|2x+6\right|\)
Vì \(\left|2x+6\right|\ge0\) nên \(A\ge4\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x+4\le0\\2x+6=0\\2x+8\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\le-4\\2x=-6\\2x\ge-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\x=-3\\x\ge-4\end{matrix}\right.\)
Vậy \(x=-3\)
TM
2
28 tháng 7 2019
a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)
28 tháng 7 2019
b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy : \(x\in\left\{0,1,2\right\}\)
Chúc học tốt nhé !!
7 tháng 3 2022
1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)
=>18x-6+18x-6=4-12x
=>36x-12=4-12x
=>48x=16
hay x=1/3
2: \(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
=>(2x-1)(3x-4)=0
=>x=1/2 hoặc x=4/3
1 tháng 9 2016
Ta có: \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
=> \(\left(2x-1\right)\in\left\{1;-1;0\right\}\)
* Nếu 2x - 1 = 1
=> 2x = 2
=> x = 2 : 2 = 1
* Nếu 2x - 1 = -1
=> 2x = (-1) + 1
=> 2x = 0
=> x = 0 : 2 = 0
* Nếu 2x - 1 = 0
=> 2x = 0 + 1
=> 2x = 1
=> x = 1 : 2
=> x = 1/2
Vậy x = { 1; 0 ; 1/2 } thì \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
CHÚC BẠN HỌC TỐT
1 tháng 9 2016
( 2x - 1 )6 = ( 2x - 1 )8
( 2x - 1 )8 - ( 2x - 1 )6 = 0
( 2x - 1 )6 . ( ( 2x - 1 )2 - 1 ) ) = 0
Vậy ( 2x - 1 )6 = 0 hoặc ( 2x - 1 )2 - 1 = 0
2x - 1 = 0 hoặc \(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}}\)
x=1/2 hoặc \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy x \(\in\){ 1/2; 0 ;1 }
Theo đầu bài ta có:
\(\left(2x-1\right)^8=\left(2x-1\right)^6\)
\(\Rightarrow2x-1=\left\{-1;0;1\right\}\)
\(\Rightarrow x=\left\{0;\frac{1}{2};1\right\}\)
\(\left(2x-1\right)^8=\left(2x-1\right)^6\)
\(\Rightarrow\left(2x-1\right)^2=2x-1\)
\(\Rightarrow4x^2-4x+1=2x+1\)
\(\Rightarrow4x^2-6x=0\)
\(\Rightarrow2x\left(2x-3\right)=0\)
\(\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1,5\end{cases}}\)
Vậy \(x=\orbr{\begin{cases}0\\1,5\end{cases}}\)