Question

Tính \(\left|-1+\left|3x-2\right|\right|>4\)

Answer 1

Ta có : \(\left|\left|3x-2\right|-1\right|>4\)

\(\Leftrightarrow\left[{}\begin{matrix}-1+\left|3x-2\right|>4\\-1+\left|3x-2\right|< -4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|3x-2\right|>5\\\left|3x-2\right|< -3\left(VL\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|3x-2\right|>5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2>5\\3x-2< -5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{7}{3}\\x< -1\end{matrix}\right.\)

Vậy.  ....

Answer 2

TH1: `-1+|3x-2| >=0 <=> |3x-2| >=1`

TH1.1: `3x-2>=0 <=> x >= 2/3`

`-1+3x-2>4`

`<=>3x>7`

`<=>x>7/3`

Vậy `x>7/3`

TH1.2: `3x-2<0 <=> x <3/2`

`-1-3x+2>4`

`<=> -3x>3`

`<=>x<-1`

Vậy `x<-1`

TH2: `-1+|3x-2|<0 <=> |3x-2|<1`

TH2.1: `3x-2>=0 <=> x >= 2/3`

`1-3x+2>4`

`<=>-3x>1`

`<=>x<-1/3`

`=>` VN.

TH2.2: `3x-2<0 <=> x<2/3`

`1+3x-2>4`

`<=>2x>5`

`<=>x>5/2`

`=>` VN.

Vậy `S=(-∞ ; -1) \vee (7/3 ; + ∞)`.