Question

Cứu≥﹏≤

Answer 1

`(1/2 x-3)^2=(1/2x-3)^4`

`(1/2x-3)^2-(1/2x-3)^4=0`

`(1/2x-3)^2 [1-(1/2x-3)^2]=0`

`[(1/2x-3=0),((1/2x-3)^2=1^2=(-1)^2):}`

`[(x=6),(1/2x-3=1),(1/2x-3=-1):}`

`[(x=6),(x=8),(x=4):}`

 

Answer 2

c) Ta có: \(\left(\dfrac{1}{2}x-3\right)^2=\left(\dfrac{1}{2}x-3\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-3\right)^2\cdot\left[\left(\dfrac{1}{2}x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{1}{2}x-3\right)^2=0\\\left(\dfrac{1}{2}x-3\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\\dfrac{1}{2}x-3=1\\\dfrac{1}{2}x-3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3\\\dfrac{1}{2}x=4\\\dfrac{1}{2}x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=8\\x=4\end{matrix}\right.\)