Question

a) x\(^2\)-3x=0

b) |x-1|=3

c) |2x-1|=1

 

Answer 1

a) Ta có: \(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: S={0;3}

b) Ta có: \(\left|x-1\right|=3\)

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

Vậy: S={4;-2}

c) Ta có: \(\left|2x-1\right|=1\)

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

Vậy: S={0;1}

Answer 2

Answer 3

a)  x² - 3x = 0

=> x ( x-3 ) = 0

=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

b)  |x-1|=3

=> \(\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)

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

c) |2x-1|=1

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

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