a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)
vậy phương trình vô nghiệm
a) (x-1)2 = 0
<=> x-1 = 0
<=> x = 1
b) (x-2)2 - 1 = 0
<=> (x-2)2 = 1
<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) (2x-1)3 = -8
<=> (2x-1)3 = -23
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(-\dfrac{1}{2}\)
d) (x+2)2 + 1 = 0
<=> (x+2)2 = -1
<=> x+2 = -1
<=> x = -3
a, \(\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2=0^2\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ......
b, \(\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy .....
c, \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy ...
d, \(\left(x+2\right)^2+1=0\)
\(\Leftrightarrow\left(x+2\right)^2=-1\)
\(\Leftrightarrow\) ko tìm dc giá trị của x thỏa mãn (do \(\left(x+2\right)^2\ge0\))
