Question

Bài 1 :tìm x:

a)[x+1/2]=3

b)[2x-1]=-4

c)[3x-1]=0

d)[2-x].[3+x]=0

Answer 1

\(a)\left|x+\dfrac{1}{2}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=3\\x+\dfrac{1}{2}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{1}{2}\\x=-3-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)

\(b)\left|2x-1\right|=-4\)

\(\Rightarrow\left|2x-1\right|=-4\)

\(\Rightarrow\left|x-1\right|=-4:2\)

\(\Rightarrow\left|x-1\right|=-2\)

Ta có: \(\left|x-1\right|\ge0\) mà \(-2< 0\)

Vậy ko có giá trị thỏa mãn đề bài.

Answer 2

[] là giá trị tuyệt đối hay là gì

Answer 3

\(\left|3x-1\right|=0\Rightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)