Question

\[\left|\frac1{h-1}\right|<1\]

Answer 1

\[(\frac{1}{h-1})^2<1^2 \\ \frac{1}{h^2-2h+1}<1 \\ h^2-2h+1>1 \\ h^2-2h>0 \\ h(h-2)>0\]

h<0 h>2

Answer 2

How do you think?

Answer 3

\[\left|\frac{1}{h-1}\right|<1\]\[\frac{1}{|h-1|}<1\]\[|h-1|>1\]this gives us Either \[h-1>1\]Or\[h-1<-1\]So, h >2 or h <0 \[\[h \epsilon (- \infty,0)U(2,\infty)\]

Answer 4

\[-1<\ \frac{ 1 }{ h-1 } <1\]

i suppose the equation proceeds from this step.

Answer 5

because we know when modulus of any equation is less then a real number then the equation can be either less then the positive value of the real number or greater then the negative value of the real number

Answer 6

Thanks @Calculator, @akitav, @praxer .
I see where i was going wrong now .

Answer 7

Differnt approaches :)