方程lgx平方-lg(x+2)=0的解集是多少?

分类: 作业答案 • 2021-12-19 16:47:28

问题描述：

答

lgx-lg(x+2) =lg[x/(x+2)]=0 所以只需x/(x+2)=1即可所以x=x+2 x-x-2=0 (x+1)(x-2)=0 x=-1或x=2 望采纳，谢谢

答

lgx平方-lg(x+2)=0 ==> lg [x^2 / (x+2)] = 0 ==> x^2 / (x +2) = 1 ==> x^2 = x +2 ==> x^2 - x -2 = 0 ==> (x -2)(x +1) = 0 ==> x + 1 = 0 ==> x =2,-1