1. ## Inequalities

Can anybody help me to solve this?

1. $\displaystyle \frac{6}{\mid x \mid + 1} < \mid x \mid$

2. $\displaystyle \frac{6}{\mid x \mid} < \mid 2x-1 \mid$

3. $\displaystyle \mid x \mid < \mid x^2-7x+6 \mid$

1. $\displaystyle x<-2, x>2$

2. $\displaystyle \{x:x<-\frac{3}{2}\} \cup \{x:x>2\}$

3. $\displaystyle x<4-\sqrt{10}, 3-\sqrt{3}<x<3+\sqrt{3}, x>4+\sqrt{10}$

2. Originally Posted by cloud5
Can anybody help me to solve this?

1. $\displaystyle \frac{6}{\mid x \mid + 1} < \mid x \mid$

2. $\displaystyle \frac{6}{\mid x \mid} < \mid 2x-1 \mid$

1. $\displaystyle x<-2, x>2$
2. $\displaystyle \{x:x<-\frac{3}{2}\} \cup \{x:x>2\}$
1. $\displaystyle \frac{6}{\mid x \mid + 1} < \mid x \mid$ or $\displaystyle 0 < \mid x \mid^2 + \mid x \mid - \,6\;\;\;\Rightarrow\;\;\; 0 < \left( \mid x \mid + 3\right)\left( \mid x \mid -2\right)$ which only happens if $\displaystyle \mid x \mid < -3$ (never) or $\displaystyle \mid x \mid > 2$, your answer.
2. $\displaystyle \frac{6}{\mid x \mid} < \mid 2x-1 \mid$ or $\displaystyle \mid 2x-1 \mid \mid x \mid > 6$ or $\displaystyle \mid 2x^2 - x \mid > 6$. Solutions of this are $\displaystyle 2x^2-x > 6$ or $\displaystyle 2x^2-x <-6$ (never). For the first $\displaystyle 2x^2-x > 6$ or $\displaystyle (2x+3)(x-2) > 0$ which gives $\displaystyle x < - \frac{3}{2}$ and $\displaystyle x > 2$.