Question

4．解不等式：|$\frac{x-1}{2x-3}$-1|＜2．

Answer 1

分析 不等式等價(jià)于|$\frac{x-2}{2x-3}$|＜2，等價(jià)于$\left\{\begin{array}{l}{\frac{x-2}{2x-3}＞-2}\\{\frac{x-2}{2x-3}＜2}\end{array}\right.$，等價(jià)于$\left\{\begin{array}{l}{\frac{5x-8}{2x-3}＞0}\\{\frac{3x-4}{2x-3}＞0}\end{array}\right.$，由此求得x的范圍．

解答 解：|$\frac{x-1}{2x-3}$-1|＜2，等價(jià)于|$\frac{x-2}{2x-3}$|＜2，等價(jià)于$\left\{\begin{array}{l}{\frac{x-2}{2x-3}＞-2}\\{\frac{x-2}{2x-3}＜2}\end{array}\right.$，等價(jià)于$\left\{\begin{array}{l}{\frac{5x-8}{2x-3}＞0}\\{\frac{3x-4}{2x-3}＞0}\end{array}\right.$，即 $\left\{\begin{array}{l}{x＜\frac{3}{2}，或x＞\frac{8}{5}}\\{x＜\frac{4}{3}，或x＞\frac{3}{2}}\end{array}\right.$．
求得它的解集為[x|x＜$\frac{4}{3}$，或x＞$\frac{8}{5}$}．

點(diǎn)評(píng) 本題主要考查絕對(duì)值不等式的解法，體現(xiàn)了轉(zhuǎn)化的數(shù)學(xué)思想，屬于基礎(chǔ)題．