1+log(3)[(x+1)/3]=log(3)3+log(3)[(x+1)/3]=log(3)(x+1)
∴log(3)(x^2-2x-3)＞log(3)(x+1)
∴x^2-2x-3＞x+1
∴x^2-3x-4＞0
问一道高一解方程红笔圈的那x^2-3x-4＞0怎么得来的
1+log(3)[(x+1)/3]=log(3)3+log(3)[(x+1)/3]=log(3)(x+1)
∴log(3)(x^2-2x-3)＞log(3)(x+1)
∴x^2-2x-3＞x+1
∴x^2-3x-4＞0