I can't solve $x-2<\frac{1}{x-1}$ We have to solve this inequation. 
$x-2<\frac{1}{x-1}$.
So we try $(x-1)(x-2)<1$ what's wrong ? 
Please, give me a hint.
 A: Yes, it's wrong because $x-1$ can be negative. 
We need
$$x-2-\frac{1}{x-1}<0$$ or
$$\frac{x^2-3x+1}{x-1}<0$$ or
$$\frac{\left(x-\frac{3+\sqrt5}{2}\right)\left(x-\frac{3-\sqrt5}{2}\right)}{x-1}<0,$$
which gives the answer.
I got $$\left(-\infty,\frac{3-\sqrt5}{2}\right)\cup\left(1,\frac{3+\sqrt5}{2}\right).$$
A: $$x-2<\frac { 1 }{ x-1 } \\ x-2-\frac { 1 }{ x-1 } <0\\ \frac { { x }^{ 2 }-3x+1 }{ x-1 } <0\\ \frac { \left( x-\frac { 3-\sqrt { 5 }  }{ 2 }  \right) \left( x-\frac { 3+\sqrt { 5 }  }{ 2 }  \right) \left( x-1 \right)  }{ { \left( x-1 \right)  }^{ 2 } } <0\\ \left( x-\frac { 3-\sqrt { 5 }  }{ 2 }  \right) \left( x-\frac { 3+\sqrt { 5 }  }{ 2 }  \right) \left( x-1 \right) <0\\ x\in \left( -\infty ,\frac { 3-\sqrt { 5 }  }{ 2 }  \right) \cup \left( 1,\frac { 3+\sqrt { 5 }  }{ 2 }  \right) \\ $$
A: hint: solve 
$$x^2-3x+2<1 , \hspace{10pt}  x>1 $$ 
$$x^2-3x+1<0 , \hspace{10pt}  x>1 $$
and 
$$x^2-3x+2>1 , \hspace{10pt}  x>1 $$ 
$$x^2-3x+1>0 , \hspace{10pt}  x<1 $$
A: If you have $x<y$ and multiply it by positive number $a$, then it is still true that $ax < ay$. If you multiply it by negative number $b$, then the inequality changes direction, so $bx > by$ is true instead.
For example, we know that $2<3$ and if we multiply it by $2$, we get $4<6$, which is still true, but if we multiply it by $-2$, it is not true that $-4<-6$, but on the contrary, $-4>-6$.
What does it have to do with solving inequalities? 
You have $$x-2 < \frac 1{x-1}$$ and want to multiplity it by $x-1$. So, what is true: $(x-2)(x-1)<1$ or $(x-2)(x-1)>1$? According to the above discussion, if $x-1>0$, the former is true, and if $x-1<0$ the latter is true. But, the problem is that we don't know whether $x-1$ is positive or negative, so we would entangle ourselves into some casework.
However, there is a common trick to avoid this, multiply by $(x-1)^2$ instead of $x-1$! We do know that $(x-1)^2>0$, so the inequality becomes $$(x-2)(x-1)^2<x-1.$$ Can you finish from here?
A: $x-1<1/(x-1) +1;$
Set $a:= x-1$, $a \not =0$.
$a < 1/a +1;$
1) Let $a >0$; then
$a^2-a -1 <0.$
$(a-1/2)^2 \lt 5/4;$
$|a-1/2| \lt (1/2)√5;$
We get for $a >0$: $a < 1/2 +(1/2)√5$.;
2) $a < 0$:
$a^2-a-1 >0;$
$(a-1/2)^2 >5/4.$
$|a-1/2| >(1/2)√5$;
We get for a <0: $1/2-a >(1/2)√5$, or
$a < 1/2-(1/2)√5$.
Combining:
$a \in (-\infty, \dfrac{1-√5}{2})\cup (0,\dfrac{1+√5}{2}).$
Now revert to $x$.
