NT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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 \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
MH
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}
DK
1
LQ
2
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 ...............
TP
Thảo Phương
CTVVIP
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 !!!
TT
5
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
hoc moi lop 5