解不等式(x-5)/(x^2-2x-3)≤1
1个回答

(x-5)/(x^2-2x-3)≤1

[(x-5)-(x^2-2x-3)]/(x^2-2x-3)≤0

(x^2-3x+2)/(x^2-2x-3)≥0

[(x-2)(x-1)]/[(x-3)(x+1)]≥0

可以转化为(x-1)(x-2)(x-3)(x+1)≥0(且x≠-1,3)

所以解为(负无穷,-1),[1,2],(3,正无穷)