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 ............
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
a, (2x-4)4 = 81 => 81 = 34 <=>2x-4 = 3 <=> 2x = 4+3 = 7 <=> x = \(\dfrac{7}{2}\)
b, (x-1)5 = -32 => -32 = (-2)5 <=> x-1 = -2 <=> x = -1
c, (2x-1)6 = (2x-1)8 <=> (2x-1)6 - (2x-1)8 <=> (2x-1)6 . [1-(2x-1)2] = 0
<=> (2x-1)6 . [(1-2x+1) . (1+2x-1)] = 0 <=> (2x-1)6 . [(2-2x) . 2x] = 0
Có:
* (2x-1)6 = 0 <=> 2x-1=0 <=> x=\(\dfrac{1}{2}\)
* 2-2x=0 <=> 2x=2 <=> x=1
* 2x=0 <=> x=0 <=> x=0
