" (ii) "(|x|-1)/(|x|-2)>=0,x in R,x!=+-2

Solve: (|x|-1)/(|x|-2)>=0,x in R,x!=+-2

(|x|-1)/(|x|-2)>=0 , x in bbb"R" , x != 2

Solve: (|x|-1)/(|x|-2)geq0,x in R ,x!=+-2.

Solve (|x|-1)/(|x|-2)>=0;x varepsilon R;x=+-2

If f: R->R is defined by f(x)={(x+2)/(x^2+3x+2) if x in R-{-1,-2}, -1 if x = -2 and 0 if x=-1. ifx=-2 then is continuous on the set

If f : R rarr R is defined by f(x) = {((x+2)/(x^(2)+3x+2), x in R-{-1,-2}),(-1, x=-2),(0, x = -1):} then f is continuous on the set

The root of (x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0, a in R are always

The root of (x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0, a in R are always

Write the discriminant of the following quadratic equations: (x-1)(2x-1)=0 (ii) x^2-2x+k=0,\ \ k in R