Question

16．不等式組$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞|\frac{2-x}{2+x}|}\end{array}\right.$的解集是（　�。�

• A． {x|0＜x＜2}
• B． {x|0＜x＜2.5}
• C． {x|0＜x＜$\sqrt{6}$}
• D． {x|0＜x＜3}

Answer 1

分析 原不等式組可化為$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞0}\\{\frac{x-3}{3+x}＜\frac{2-x}{2+x}＜\frac{3-x}{3+x}}\end{array}\right.$，解不等式組可得．

解答 解：原不等式組可化為$\left\{\begin{array}{l}{x＞0}\\{\frac{3-x}{3+x}＞0}\\{\frac{x-3}{3+x}＜\frac{2-x}{2+x}＜\frac{3-x}{3+x}}\end{array}\right.$，
解$\frac{3-x}{3+x}$＞0可得-3＜x＜3，結(jié)合x(chóng)＞0可得0＜x＜3，
解不等式$\frac{2-x}{2+x}$＜$\frac{3-x}{3+x}$可得-3＜x＜-2或x＞0，
解不等式$\frac{2-x}{2+x}$＞$\frac{x-3}{3+x}$可得-3＜x＜-$\sqrt{6}$或-2＜x＜$\sqrt{6}$
綜合可得得0＜x＜$\sqrt{6}$．
故選：C．

點(diǎn)評(píng) 本題考查分式不等式的解集，涉及穿根法解不等式，屬中檔題．