\(a.\left|x\right|< 5\Leftrightarrow-5< x< 5\)

\(b.\left|2x+1\right|< 3\Leftrightarrow\left\{{}\begin{matrix}-3< 2x+1\\2x+1< 3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x< 1\end{matrix}\right.\)\(\Leftrightarrow-2< x< 1\)

\(c.\left|x-1\right|< 2\Leftrightarrow\left\{{}\begin{matrix}x-1>-2\\x-1< 2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x< 3\end{matrix}\right.\)\(\Leftrightarrow-1< x< 3\)

help meeeeeeeeeeeeeee

a: |x|<5

=>-5<x<5

b: =>-3<2x+1<3

=>-4<2x<2

=>-2<x<1

c: =>-2<x-1<2

=>-1<x<3

a: |x|<5

=>-5<x<5

b: =>-3<2x+1<3

=>-4<2x<2

=>-2<x<1

c: =>-2<x-1<2

=>-1<x<3

aaaaaaaaaaaaaaaaaaaaa

giúp mình với mình cần gấp lắmmmm

a: |x|<5

=>-5<x<5

b: =>-3<2x+1<3

=>-4<2x<2

=>-2<x<1

c: =>-2<x-1<2

=>-1<x<3

\(A=2\left|3x-2\right|-1\ge-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow3x-2=0\Leftrightarrow x=\frac{2}{3}\)

\(B=5\left|1-4x\right|-1\ge-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow1-4x=0\Leftrightarrow x=\frac{1}{4}\)

\(A=5-\left|2x-1\right|\le5\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

\(\left|2x-1\right|=\left|2x+3\right|\)

\(\Leftrightarrow\left|2x-1\right|^2=\left|2x+3\right|^2\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(2x+3\right)^2\)  (\(\left|a\right|^2=a^2\) )

\(\Leftrightarrow\left(2a-1\right)^2-\left(2a+3\right)^2=0\)

\(\Leftrightarrow\left(2a-1-2a-3\right)\left(2a-1+2a+3\right)=0\)

\(\Leftrightarrow-4\left(4a+2\right)=0\)

\(\Rightarrow x=-\frac{1}{2}\)

Vậy ............

cái chỗ ( \(\left|a\right|^2=a^2\) ) là lưu ý nha