a, (2x-1)^4=81
=> (2x-1)^4= 3^4
=>2x-1=3
=>2x=3+1=4
=>x=4/2=2
b, (x-1)^5=-32
=>(x-1)^5=(-2)^5
=>x-1=-2
=>x=(-2)+1
=>x=-1
a) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=\left(\pm3\right)^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy............
b) \(\left(x-1\right)^5=-32\)
\(\Rightarrow\left(x-1\right)^5=-2^5\)
\(\Rightarrow x-1=-2\Rightarrow x=-1\)
Vậy.......
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Rightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy x = 1/2; x = 1; x = 0
a) \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left(2x-1\right)^4=3^4\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy ............................
b) \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-2+1\)
\(\Leftrightarrow x=-1\)
1.
a) \(\left(2x-1\right)^4=81\)
\(\left(2x-1\right)^4=\left(\pm3\right)^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
b) \(\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(\Rightarrow x-1=-2\Rightarrow x=-3\)
Vậy...
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left(2x-1\right)^2=\left(\pm1\right)^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=x=0\end{matrix}\right.\)
Vậy...
Tik mik nha !!!
