Question

19．若规定$|\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}|$=ad-bc，则$|\begin{array}{l}{1}&{2}\\{x}&{{x}^{2}}\end{array}|$＜3的解集是{x|-1＜x＜3}．

Answer 1

分析 由已知得x²-2x＜3，由此能求出$|\begin{array}{l}{1}&{2}\\{x}&{{x}^{2}}\end{array}|$＜3的解集．

解答 解：∵规定$|\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}|$=ad-bc，则$|\begin{array}{l}{1}&{2}\\{x}&{{x}^{2}}\end{array}|$＜3，
∴x²-2x＜3，
解得-1＜x＜3．
∴$|\begin{array}{l}{1}&{2}\\{x}&{{x}^{2}}\end{array}|$＜3的解集是{x|-1＜x＜3}．
故答案为：{x|-1＜x＜3}．

点评 本题考查不等的解集的求法，是基础题，解题时要认真审题，注意二除行列式的性质的合理运用．