Với mọi \(x\) ta luôn luôn có: \(\hept{\begin{cases}\left|x-2\right|\ge0\\\left|3-2x\right|\ge0\end{cases}}\Rightarrow\left|x-2\right|+\left|3-2x\right|\ge0\Rightarrow2x+1\ge0\Rightarrow x\ge\frac{-1}{2}\)

Trường hợp 1: \(x< \frac{3}{2}\)

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

\(\Rightarrow2-x+3-2x=2x+1\)

\(\Rightarrow5-3x=2x+1\)

\(\Rightarrow-3x-2x=1-5\)

\(\Rightarrow-5x=-4\)

\(\Rightarrow x=\frac{4}{5}\) (Thoả mãn)

Trường hợp 2: \(x>2\)

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

\(\Rightarrow x-2+2x-3=2x+1\)

\(\Rightarrow3x-5=2x+1\)

\(\Rightarrow x=6\) (Thoả mãn)

Trường hợp 3: \(\frac{3}{2}\le x\le2\)

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

\(\Rightarrow2-x+2x-3=2x+1\)

\(\Rightarrow-1+x=2x+1\)

\(\Rightarrow-1-1=2x-x\)

\(\Rightarrow x=-2\) (Loại)

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

\(\Rightarrow\orbr{\begin{cases}x=2x-1\\x=1-2x\end{cases}}\Rightarrow\orbr{\begin{cases}-x=-1\\3x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\)

đây nhé

đây nhé

a) \(\frac{-18}{91}\) và \(\frac{-23}{114}\)

\(\frac{-18}{91}=\frac{-18.114}{91.114}=\frac{-2052}{\text{10374}}\) 

\(\frac{-23}{114}=\frac{-23.91}{114.91}=\frac{\text{-2093}}{10374}\)

Ta có:

\(\frac{-2052}{10374}< \frac{-2093}{10374}\)

\(\Rightarrow\)\(\frac{-18}{91}< \frac{-23}{114}\)

b) \(\frac{-22}{35}\) và \(\frac{-103}{177}\)

\(\frac{-22}{35}=\frac{-22.177}{35.177}=\frac{\text{-3894}}{\text{6195}}\)

\(\frac{-103}{177}=\frac{-103.35}{177.35}=\frac{\text{-3605}}{6195}\)

Ta có:

\(\frac{-3894}{6195}< \frac{-3605}{6195}\)

\(\Rightarrow\)\(\frac{-22}{35}< \frac{-103}{177}\)