Question

[Nguyễn Khắc Hoàng Quân]

rút gọn biểu thức:

2|x - 3| - |4x - 1|.

Answer 1

[Trí Tiên亗]

Xét \(x< \frac{1}{4}\Rightarrow\hept{\begin{cases}4x-1< 0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=1-4x\\\left|x-3\right|=3-x\end{cases}}\)

Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(1-4x\right)=2x+5\)

Xét \(\frac{1}{4}\le x< 3\Rightarrow\hept{\begin{cases}4x-1\ge0\\x-3< 0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=3-x\end{cases}}\)

Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(3-x\right)-\left(4x-1\right)=-6x+7\)

\(x\ge3\Rightarrow\hept{\begin{cases}4x-1>0\\x-3\ge0\end{cases}\Rightarrow}\hept{\begin{cases}\left|4x-1\right|=4x-1\\\left|x-3\right|=x-3\end{cases}}\)

Khi đó biểu thức : \(2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)=-2x-5\)

Answer 2

[Nobi Nobita]

TH1: Nếu \(x< \frac{1}{4}\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=-\left(4x-1\right)\end{cases}}\)

\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)+\left(4x-1\right)\)

\(=-2x+6+4x-1=2x+5\)

TH2: Nếu \(\frac{1}{4}\le x\le3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=-\left(x-3\right)\\\left|4x-1\right|=4x-1\end{cases}}\)

\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=-2\left(x-3\right)-\left(4x-1\right)\)

\(=-2x+6-4x+1=-6x+7\)

TH3: Nếu \(x>3\)\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=x-3\\\left|4x-1\right|=4x-1\end{cases}}\)

\(\Rightarrow2\left|x-3\right|-\left|4x-1\right|=2\left(x-3\right)-\left(4x-1\right)\)

\(=2x-6-4x+1=-2x-5\)