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25 tháng 9 2017

\(\left(2x-4\right)^4=81\)

\(\left(2x-4\right)^4=3^4\)

\(\Rightarrow2x-4=3\)

\(\Rightarrow2x=7\)

\(\Rightarrow x=\frac{7}{2}\)

vay \(x=\frac{7}{2}\)

\(\left(x-1\right)^5=-32\)

\(\left(x-1\right)^5=\left(-2\right)^5\)

\(\Rightarrow x-1=-2\)

\(\Rightarrow x=-1\)

vay \(x=-1\)

\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)

\(\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)

\(\left(2x-1\right)^6\left(1-2x+1\right)\left(1+2x-1\right)=0\)

\(\left(2x-1\right)^6\left(-2x+2\right)\left(2x\right)=0\)

\(\Rightarrow\left(2x-1\right)^6=0\)hoac \(\Rightarrow\orbr{\begin{cases}-2x+2=0\\2x=0\end{cases}}\)

\(\Rightarrow2x-1=0\) hoac \(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

\(\Rightarrow x=\frac{1}{2}\)hoac \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

14 tháng 7 2015

a. (2x-1)4=81

=>(2x-1)4=34

=>2x-1=3

=>2x=3+1

=>2x=4

=>x=4:2

=>x=2

b.(x-1)5=-32

=>(x-1)5=(-2)5

=>x-1=-2

=>x=-2+1

=>x=-1

c.(2x-1)6=(2x-1)8

mà chỉ có: (-1)6=(-1)8; 06=08; 16=18

=> để (2x-1) \(\in\){-1;0;1} thì x \(\in\){0; 1/2; 1}

25 tháng 10 2017

(2x-1)^4=3^4

=>2x-1=3

=>2x=4

=>x=2

19 tháng 7 2016

minh ko bit

20 tháng 7 2017

a ) \(\left(2x-1\right)^4=81\)

\(\Leftrightarrow\left(2x-1\right)^4=3^4\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow x=2\)

Vậy \(x=2.\)

b ) \(\left(x-1\right)^5=-32\)

\(\Leftrightarrow\) \(\left(x-1\right)^5=-2^5\)

\(\Leftrightarrow x-1=-2\)

\(\Leftrightarrow x=-1\)

Vậy \(x=-1.\)

c ) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[\left(1-2x+1\right)\left(1+2x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2-2x\right).2x\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\2-2x=0\\2x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)

Vậy ...............

20 tháng 7 2017

a) (2x-1)4​=81

\(\Leftrightarrow\)\(\left[\begin{array}{} (2x-1)^4=(3)^4\\ (2x-1)^4=(-3)^4 \end{array}\right.\)

\(\Rightarrow\)\(\left[\begin{array}{} 2x-1=3\\ 2x-1=-3 \end{array}\right.\)

\(\Rightarrow\)\(\left[\begin{array}{} 2x=3+1\\ 2x=-3+1 \end{array}\right.\)

\(\Rightarrow\)\(\left[\begin{array}{} 2x=4\\ 2x=-2 \end{array}\right.\)

\(\Rightarrow\)\(\left[\begin{array}{} x=4:2\\ x=-2:2 \end{array}\right.\)

\(\Rightarrow\)\(\left[\begin{array}{} x=2\\ x=-1 \end{array}\right.\)

Vậy x=2 hoặc x=-1

b) (x-1)5​= -32

\(\Leftrightarrow\)\( (x-1)^5=(-2)^5 \)

\(\Rightarrow\)\( (x-1)=-2 \)

\(\Rightarrow\)\( x=-2+1 \)

\(\Rightarrow\)\( x=-1 \)

Vậy x=-1

c) ( 2x-1)6​= ( 2x-1)8

\(\Leftrightarrow\) (2x-1)6=(2x-1)8.

\(\Leftrightarrow\)(2x-1)8-(2x-1)6=0.

\(\Leftrightarrow\)(2x-1)6)[(2x-1)2-1]=0.

\(\Leftrightarrow\)(2x-1)6(2x-1+1)(2x+1+1)=0.

\(\Leftrightarrow\)(2x-1)62x(2x+2)=0.

\(\Leftrightarrow\)(2x-1)6<=>2x(2x-1)=0.\(\Rightarrow x=\dfrac{1}{2}\)

hoặc 2x=0\(\Rightarrow\)x=0

hoặc 2x+2=0\(\Rightarrow\)2x=-2\(\Leftrightarrow\)x=-2:2\(\Leftrightarrow\)x=-1

Vậy x=\(\dfrac{1}{2}\)hoặc x=0 hoặc x=-1

Chúc bạn học tốt !!!

19 tháng 9 2017

a/ \(\left(2x-4\right)^4=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-4\right)^4=3^4\\\left(2x-4\right)^4=\left(-3\right)^4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=3\\2x-4=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy .......

b/ \(\left(x-1\right)^5=-32\)

\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)

\(\Leftrightarrow x-1=-2\)

\(\Leftrightarrow x=-1\)

Vậy ............

19 tháng 9 2017

a) \(\left(2x-4\right)^4=81\\ \left(2x-4\right)^4=3^4\\\Rightarrow2x-4=3\\ 2x=3+4\\ 2x=7\\ x=7:2\\ x=\dfrac{7}{2} \)

Vậy \(x=\dfrac{7}{2}\)

b) \(\left(x-1\right)^5=-32\\ \left(x-1\right)^5=\left(-2\right)^5\\ \Rightarrow x-1=-2\\ x=-2+1\\ x=-1\)

Vậy \(x=-1\)

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \)

Suy ra không tìm được x