Question

Finden Sie alle x ∈ ℝ

$\frac{1}{(x+2)^{2}} \leq \frac{1}{x-1}$

$\left|\frac{3 x+4}{x-2}\right| \leq 1$

Answer 1

1. Aufgabe:

$$\frac { 1 }{ { \left( x+2 \right) }^{ 2 } } \le \frac { 1 }{ x-1 }$$$$\Leftrightarrow 1\le \frac { { \left( x+2 \right) }^{ 2 } }{ x-1 }$$$$Fall1:$$$$x-1<0\Leftrightarrow x<1$$$$\Rightarrow 1\le \frac { { \left( x+2 \right) }^{ 2 } }{ x-1 }$$$$\Leftrightarrow x-1\ge { \left( x+2 \right) }^{ 2 }={ x }^{ 2 }+4x+4$$$$\Leftrightarrow { x }^{ 2 }+3x+5\le 0$$$$\Leftrightarrow { x }^{ 2 }+3x\le -5$$$$\Leftrightarrow { x }^{ 2 }+3x+{ 1,5 }^{ 2 }\le { 1,5 }^{ 2 }-5=-2,75$$$$\Leftrightarrow { (x+1,5) }^{ 2 }\le =-2,75$$$$\Rightarrow L_{ 1 }=\left\{ \quad \right\}$$$$Fall2:$$$$x-1>0\Leftrightarrow x>1$$$$\Rightarrow 1\le \frac { { \left( x+2 \right) }^{ 2 } }{ x-1 }$$$$\Leftrightarrow x-1\le { \left( x+2 \right) }^{ 2 }={ x }^{ 2 }+4x+4$$$$\Leftrightarrow { x }^{ 2 }+3x+5\ge 0$$$$\Leftrightarrow { x }^{ 2 }+3x\ge -5$$$$\Leftrightarrow { x }^{ 2 }+3x+{ 1,5 }^{ 2 }\ge { 1,5 }^{ 2 }-5=-2,75$$$$\Leftrightarrow { (x+1,5) }^{ 2 }\ge =-2,75$$$$\Leftrightarrow { (x+1,5) }^{ 2 }\ge =0$$$$\Leftrightarrow { x }\ge =-1,5$$$$\Rightarrow { L }_{ 2 }=\left\{ x\in R|x>1\wedge x\ge =-1,5 \right\} =\left\{ x\in R|x>1 \right\}$$$$\Rightarrow L={ L }_{ 1 }\cup { L }_{ 2 }=\left\{ \quad \right\} \cup \left\{ x\in R|x>1 \right\} =\left\{ x\in R|x>1 \right\}$$

2, Aufgabe:

$$\left| \frac { 3x+4 }{ x-2 } \right| \le 1$$$$Fall1:\left| \frac { 3x+4 }{ x-2 } \right| <0$$$$\Leftrightarrow (3x+4<0\wedge x-2>0)\vee (3x+4>0\wedge x-2<0)$$$$\Leftrightarrow (3x<-4\wedge x>2)\vee (3x>-4\wedge x<2)$$$$\Leftrightarrow falsch\vee (x>-\frac { 4 }{ 3 } \wedge x<2)$$$$\Leftrightarrow -\frac { 4 }{ 3 } <x<2$$$$\Rightarrow \left| \frac { 3x+4 }{ x-2 } \right| \le 1$$$$\Leftrightarrow \frac { -3x-4 }{ x-2 } \le 1$$$$\Leftrightarrow -3x-4\ge x-2$$$$\Leftrightarrow -4x\ge 2$$$$\Leftrightarrow x\le -\frac { 1 }{ 2 }$$$$\Rightarrow { L }_{ 1 }=\left\{ x\in R|-\frac { 4 }{ 3 } <x<2\wedge x\le -\frac { 1 }{ 2 } \right\} =\left\{ x\in R|-\frac { 4 }{ 3 } <x\le -\frac { 1 }{ 2 } \right\}$$$$Fall2:\left| \frac { 3x+4 }{ x-2 } \right| \ge 0$$$$\Leftrightarrow (3x+4\ge 0\wedge x-2>0)\vee (3x+4\le 0\wedge x-2<0)$$$$\Leftrightarrow (3x\ge -4\wedge x>2)\vee (3x\le -4\wedge x<2)$$$$\Leftrightarrow (x\ge -\frac { 4 }{ 3 } \wedge x>2)\vee (x\le -\frac { 4 }{ 3 } \wedge x<2)$$$$\Leftrightarrow (x>2)\vee (x\le -\frac { 4 }{ 3 } )$$$$\Rightarrow \left| \frac { 3x+4 }{ x-2 } \right| \le 1$$$$\Leftrightarrow \frac { 3x+4 }{ x-2 } \le 1$$$$Fall 2a:x>2$$$$\Rightarrow \frac { 3x+4 }{ x-2 } \le 1$$$$\Leftrightarrow 3x+4\le x-2$$$$\Leftrightarrow 2x\le -6$$$$\Leftrightarrow x\le -3$$$$\Rightarrow L_{ 2a }=\left\{ x\in R|x>2\wedge x<-3 \right\} =\left\{ \quad \right\}$$$$Fall2b:x\le -\frac { 4 }{ 3 }$$$$\Rightarrow \frac { 3x+4 }{ x-2 } \le 1$$$$\Leftrightarrow 3x+4\ge x-2$$$$\Leftrightarrow 2x\ge -6$$$$\Leftrightarrow x\ge -3$$$$\Rightarrow L_{ 2b }=\left\{ x\in R|x\le -\frac { 4 }{ 3 } \wedge x\ge -3 \right\} =\left\{ x\in R|-3\le x\le -\frac { 4 }{ 3 } \right\}$$$$\Rightarrow { L }_{ 2 }={ L }_{ 2a }\cup { L }_{ 2b }=\left\{ \quad \right\} \cup \left\{ x\in R|-3\le x\le -\frac { 4 }{ 3 } \right\} =\left\{ x\in R|-3\le x\le -\frac { 4 }{ 3 } \right\}$$$$\Rightarrow L={ L }_{ 1 }\cup { L }_{ 2 }=\left\{ x\in R|-\frac { 4 }{ 3 } <x\le -\frac { 1 }{ 2 } \right\} \cup \left\{ x\in R|-3\le x\le -\frac { 4 }{ 3 } \right\} =\left\{ x\in R|-3\le x\le -\frac { 1 }{ 2 } \right\}$$