$a\text{)}\frac{3}{\left|x-1\right|-1}=2$  

Założenia:

$\left|x-1\right|-1\ne 0$  

$\left|x-1\right|\ne 1$  

Zatem:

$x-1\ne 1\wedge x-1\ne -1$  

$x\ne 2\wedge x\ne 0$  

$D=\mathbb{R}-\left\{0,2\right\}$  

Rozwiążmy podane równanie:

$\frac{3}{\left|x-1\right|-1}=2$  

$\frac{3}{\left|x-1\right|-1}-2=0$  

$\frac{3}{\left|x-1\right|-1}-\frac{2\left(\left|x-1\right|-1\right)}{\left|x-1\right|-1}=0$  

$\frac{3-2\left(\left|x-1\right|-1\right)}{\left|x-1\right|-1}=0$  

$3-2\left(\left|x-1\right|-1\right)=0$  

$-2\left|x-1\right|+2=-3$  

$-2\left|x-1\right|=-5$  

$\left|x-1\right|=\frac{5}{2}$  

Więc:

$x-1=\frac{5}{2}\vee x-1=-\frac{5}{2}$  

$x=\frac{7}{2}\vee x=-\frac{3}{2}$  

$x=3\frac{1}{2}\vee x=-1\frac{1}{2}$  

Rozwiązania tego równania to x=3 ¹/₂ lub x=-1 ¹/₂.

---

$b\text{)}\frac{4}{\left|x+3\right|}=\frac{1}{\left|x-5\right|}$  

Założenia:

$\left|x+3\right|\ne 0\wedge \left|x-5\right|\ne 0$  

Zatem:

$x+3\ne 0\wedge x-5\ne 0$  

$x\ne -3\wedge x\ne 5$  

$D=\mathbb{R}-\left\{-3,5\right\}$  

Rozwiążmy podane równanie:

$\frac{4}{\left|x+3\right|}=\frac{1}{\left|x-5\right|}$  

Z definicji wartości bezwzględnej wiemy, że:

$\left|x+3\right|=\{\begin{array}{l}x+3\text{jesˊli}x+3>0\\ -x-3\text{jesˊli}x+3<0\end{array}$  

$\left|x+3\right|=\{\begin{array}{l}x+3\text{jesˊli}x>-3\\ -x-3\text{jesˊli}x<-3\end{array}$  

Oraz

$\left|x-5\right|=\{\begin{array}{l}x-5\text{jesˊli}x-5>0\\ -x+5\text{jesˊli}x-5<0\end{array}$  

$\left|x+3\right|=\{\begin{array}{l}x-5\text{jesˊli}x>5\\ -x+5\text{jesˊli}x<5\end{array}$  

Rozważmy 3 przypadki:

$I.x\in \left(-\infty ,-3\right)$  

$\frac{4}{-x-3}=\frac{1}{-x+5}$  

$-\frac{4}{x+3}=-\frac{1}{x-5}$

$\frac{4}{x+3}=\frac{1}{x-5}$

$4\left(x-5\right)=x+3$  

$4x-20=x+3$  

$3x=23$  

$x=\frac{23}{3}\notin \left(-\infty ,-3\right)$  

$II.x\in \left(-3,5\right)$  

$\frac{4}{x+3}=\frac{1}{-x+5}$  

$\frac{4}{x+3}=\frac{1}{5-x}$

$4\left(5-x\right)=x+3$  

$20-4x=x+3$  

$17=5x$  

$x=\frac{17}{5}$  

$x=3\frac{2}{5}\in \left(-3,5\right)$  

$III.x\in \left(5,+\infty \right)$  

$\frac{4}{x+3}=\frac{1}{x-5}$  

$4\left(x-5\right)=x+3$  

$4x-20=x+3$  

$3x=23$  

$x=\frac{23}{3}$  

$x=7\frac{2}{3}\in \left(5,+\infty \right)$  

Rozwiązania tego równania to x=3 ²/₅ lub x=7 ²/₃.

---

$c\text{)}\frac{\left|x-1\right|}{3}-\frac{1}{\left|2x+3\right|}=0$  

Założenia:

$\left|2x+3\right|\ne 0$  

Zatem:

$2x+3\ne 0$  

$2x\ne -3$  

$x\ne -\frac{3}{2}$  

$D=\mathbb{R}-\left\{-\frac{3}{2}\right\}$  

Rozwiążmy podane równanie:

$\frac{\left|x-1\right|}{3}-\frac{1}{\left|2x+3\right|}=0$  

Z definicji wartości bezwzględnej wiemy, że:

$\left|x-1\right|=\{\begin{array}{l}x-1\text{jesˊli}x-1\ge 0\\ -x+1\text{jesˊli}x-1<0\end{array}$  

$\left|x-1\right|=\{\begin{array}{l}x-1\text{jesˊli}x\ge 1\\ 1-x\text{jesˊli}x<1\end{array}$  

Oraz

$\left|2x+3\right|=\{\begin{array}{l}2x+3\text{jesˊli}2x+3>0\\ -2x-3\text{jesˊli}2x+3<0\end{array}$  

$\left|2x+3\right|=\{\begin{array}{l}2x+3\text{jesˊli}x>-\frac{3}{2}\\ -2x-3\text{jesˊli}x<-\frac{3}{2}\end{array}$  

Rozważmy 3 przypadki:

$I.x\in \left(-\infty ,-\frac{3}{2}\right)$  

$\frac{1-x}{3}-\frac{1}{-2x-3}=0$  

$\frac{\left(1-x\right)\left(-2x-3\right)}{3\left(-2x-3\right)}-\frac{3}{3\left(-2x-3\right)}=0$  

$\frac{\left(1-x\right)\left(-2x-3\right)-3}{3\left(-2x-3\right)}=0$  

$\left(-1\right)\left(x-1\right)\left(-1\right)\left(2x+3\right)-3=0$

$\left(x-1\right)\left(2x+3\right)-3=0$

$2x^2+3x-2x-3-3=0$

$2x^2+x-6=0$

$\Delta =1^2-4\cdot 2\cdot \left(-6\right)=1+48=49,\sqrt{\Delta }=7$  

$x_1=\frac{-1-7}{2\cdot 2}=\frac{-8}{4}=-2\in \left(-\infty ,-\frac{3}{2}\right)$  

$x_2=\frac{-1+7}{2\cdot 2}=\frac{6}{4}=\frac{3}{2}\notin \left(-\infty ,-\frac{3}{2}\right)$  

$II.x\in \left(-\frac{3}{2},1\right)$  

$\frac{1-x}{3}-\frac{1}{2x+3}=0$  

$\frac{\left(1-x\right)\left(2x+3\right)}{3\left(2x+3\right)}-\frac{3}{3\left(2x+3\right)}=0$  

$\frac{\left(1-x\right)\left(2x+3\right)-3}{3\left(2x+3\right)}=0$  

$\left(1-x\right)\left(2x+3\right)-3=0$

$2x+3-2x^2-3x-3=0$

$-2x^2-x=0$

$2x^2+x=0$  

$x\left(2x+1\right)=0$  

$x=0\vee 2x+1=0$  

$x=0\in \left(-\frac{3}{2},1\right)\vee x=-\frac{1}{2}\in \left(-\frac{3}{2},1\right)$  

$III.x\in ⟨1,+\infty )$  

$\frac{x-1}{3}-\frac{1}{2x+3}=0$  

$\frac{\left(x-1\right)\left(2x+3\right)}{3\left(2x+3\right)}-\frac{3}{3\left(2x+3\right)}=0$  

$\frac{\left(x-1\right)\left(2x+3\right)-3}{3\left(2x+3\right)}=0$  

$\left(x-1\right)\left(2x+3\right)-3=0$

$2x^2+3x-2x-3-3=0$

$2x^2+x-6=0$

$\Delta =1^2-4\cdot 2\cdot \left(-6\right)=1+48=49,\sqrt{\Delta }=7$  

$x_1=\frac{-1-7}{2\cdot 2}=\frac{-8}{4}=-2\notin ⟨1,+\infty )$  

$x_2=\frac{-1+7}{2\cdot 2}=\frac{6}{4}=\frac{3}{2}=1\frac{1}{2}\in ⟨1,+\infty )$  

Rozwiązania tego równania to x=-2 lub x=0 lub  x=- ¹/₂ lub x=1¹/₂.

---

$d\text{)}\frac{x}{\left|x-1\right|}-\frac{2}{x}=\frac{5}{6}$  

Założenia:

$\left|x-1\right|\ne 0\wedge x\ne 0$  

Zatem:

$x-1\ne 0\wedge x\ne 0$  

$x\ne 1\wedge x\ne 0$  

$D=\mathbb{R}-\left\{0,1\right\}$  

Rozwiążmy podane równanie:

$\frac{x}{\left|x-1\right|}-\frac{2}{x}=\frac{5}{6}$  

Z definicji wartości bezwzględnej wiemy, że:

$\left|x-1\right|=\{\begin{array}{l}x-1\text{jesˊli}x-1>0\\ -x+1\text{jesˊli}x-1<0\end{array}$  

$\left|x-1\right|=\{\begin{array}{l}x-1\text{jesˊli}x>1\\ 1-x\text{jesˊli}x<1\end{array}$  

uwzględniając założenie x≠0 rozważmy 2 przypadki:

$I.x\in \left(-\infty ,0\right)\cup \left(0,1\right)$  

$\frac{x}{1-x}-\frac{2}{x}=\frac{5}{6}$  

$\frac{x^2-2\left(1-x\right)}{\left(1-x\right)x}=\frac{5}{6}$  

$\frac{x^2-2+2x}{\left(1-x\right)x}=\frac{5}{6}$

$6\left(x^2+2x-2\right)=5\left(1-x\right)x$  

$6x^2+12x-12=5x-5x^2$

$11x^2+7x-12=0$

$\Delta =7^2-4\cdot 11\cdot \left(-12\right)=49+528=577,\sqrt{\Delta }=\sqrt{577}$  

$x_1=\frac{-7-\sqrt{577}}{2\cdot 11}=\frac{-7-\sqrt{577}}{22}\in \left(-\infty ,0\right)\cup \left(0,1\right)$  

$x_2=\frac{-7+\sqrt{577}}{2\cdot 11}=\frac{-7+\sqrt{577}}{22}\in \left(-\infty ,0\right)\cup \left(0,1\right)$  

$II.x\in \left(1,+\infty \right)$  

$\frac{x}{x-1}-\frac{2}{x}=\frac{5}{6}$  

$\frac{x^2-2\left(x-1\right)}{\left(x-1\right)x}=\frac{5}{6}$  

$\frac{x^2-2x+2}{\left(x-1\right)x}=\frac{5}{6}$

$6\left(x^2-2x+2\right)=5\left(x-1\right)x$  

$6x^2-12x+12=5x^2-5x$

$x^2-7x+12=0$

$\Delta ={\left(-7\right)}^2-4\cdot 1\cdot 12=49-48=1,\sqrt{\Delta }=1$  

$x_1=\frac{7-1}{2}=\frac{6}{2}=3\in \left(1,+\infty \right)$  

$x_2=\frac{7+1}{2}=\frac{8}{2}=4\in \left(1,+\infty \right)$  

Rozwiązania tego równania to x=^-7-√577/₂₂ lub  x=^-7+√577/₂₂  lub  x=3 lub x=4.

---

$e\text{)}\frac{8}{\left|x\right|}+\left|x-2\right|=x$  

Założenia:

$\left|x\right|\ne 0$  

Zatem:

$x\ne 0$  

$D=\mathbb{R}-\left\{0\right\}$  

Rozwiążmy podane równanie:

$\frac{8}{\left|x\right|}+\left|x-2\right|=x$  

Z definicji wartości bezwzględnej wiemy, że:

$\left|x\right|=\{\begin{array}{l}x\text{jesˊli}x>0\\ -x+1\text{jesˊli}x<0\end{array}$  

Oraz

$\left|x-2\right|=\{\begin{array}{l}x-2\text{jesˊli}x-2\ge 0\\ -x+2\text{jesˊli}x-2<0\end{array}$  

$\left|x-2\right|=\{\begin{array}{l}x-2\text{jesˊli}x\ge 2\\ 2-x\text{jesˊli}x<2\end{array}$  

Rozważmy 3 przypadki:

$I.x\in \left(-\infty ,0\right)$  

$\frac{8}{-x}+2-x=x$  

$-\frac{8}{x}+2=2x$  

$-\frac{8}{x}+\frac{2x}{x}=2x$  

$\frac{2x-8}{x}=2x$  

$2x-8=2x^2$

$2x^2-2x+8=0$  

$\Delta ={\left(-2\right)}^2-4\cdot 2\cdot 8=4-64<0$  

$II.x\in \left(0,2\right)$  

$\frac{8}{x}+2-x=x$  

$\frac{8}{x}+2=2x$  

$\frac{8}{x}+\frac{2x}{x}=2x$  

$\frac{2x+8}{x}=2x$  

$2x+8=2x^2$

$2x^2-2x-8=0$  

$\Delta ={\left(-2\right)}^2-4\cdot 2\cdot \left(-8\right)=4+64=68,\sqrt{\Delta }=2\sqrt{17}$  

$x_1=\frac{2-2\sqrt{17}}{2\cdot 2}=\frac{1-\sqrt{17}}{2}\notin \left(0,2\right)$  

$x_2=\frac{2+2\sqrt{17}}{2\cdot 2}=\frac{1+\sqrt{17}}{2}\notin \left(0,2\right)$  

$III.x\in \left(2,+\infty \right)$  

$\frac{8}{x}+x-2=x$  

$\frac{8}{x}-2=0$  

$\frac{8}{x}=2$  

$8=2x$  

$x=4\in \left(2,+\infty \right)$

 

Jedyne rozwiązanie tego równania to  x=4.

---

$f\text{)}\frac{\left|x+2\right|}{x^2+x-2}+\frac{3}{4}=\frac{2}{\left|x+1\right|}$  

Założenia:

$x^2+x-2\ne 0\wedge \left|x+1\right|\ne 0$  

Zatem:

$x^2-x+2x-2\ne 0\wedge x+1\ne 0$  

$x\left(x-1\right)+2\left(x-1\right)\ne 0\wedge x\ne -1$  

$\left(x-1\right)\left(x+2\right)\ne 0\wedge x\ne -1$  

$x-1\ne 0\wedge x+2\ne 0\wedge x\ne -1$  

$x\ne 1\wedge x\ne -2\wedge x\ne -1$  

$D=\mathbb{R}-\left\{-2,-1,1\right\}$  

Rozwiążmy podane równanie:

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

Czyli:

$\frac{\left|x+2\right|}{\left(x-1\right)\left(x+2\right)}+\frac{3}{4}=\frac{2}{\left|x+1\right|}$  

Z definicji wartości bezwzględnej wiemy, że:

$\left|x+2\right|=\{\begin{array}{l}x+2\text{jesˊli}x+2>0\\ -x-2\text{jesˊli}x+2<0\end{array}$  

$\left|x+2\right|=\{\begin{array}{l}x+2\text{jesˊli}x>-2\\ -x-2\text{jesˊli}x<-2\end{array}$  

Oraz

$\left|x+1\right|=\{\begin{array}{l}x+1\text{jesˊli}x+1>0\\ -x-1\text{jesˊli}x+1<0\end{array}$  

$\left|x+1\right|=\{\begin{array}{l}x+1\text{jesˊli}x>-1\\ -x-1\text{jesˊli}x<-1\end{array}$  

Rozważmy 3 przypadki:

$I.x\in \left(-\infty ,-2\right)$  

$\frac{-x-2}{\left(x-1\right)\left(x+2\right)}+\frac{3}{4}=\frac{2}{-x-1}$  

$\frac{-\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{3}{4}=-\frac{2}{x+1}$  

$-\frac{1}{x-1}+\frac{3}{4}=-\frac{2}{x+1}$  

$-\frac{4}{4\left(x-1\right)}+\frac{3\left(x-1\right)}{4\left(x-1\right)}=-\frac{2}{x+1}$  

$\frac{-4+3\left(x-1\right)}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\frac{-4+3x-3}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\frac{3x-7}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\left(3x-7\right)\left(x+1\right)=4\left(x-1\right)\cdot \left(-2\right)$  

$3x^2+3x-7x-7=-8x+8$  

$3x^2+4x-15=0$  

$\Delta =4^2-4\cdot 3\cdot \left(-15\right)=16+180=196,\sqrt{\Delta }=14$  

$x_1=\frac{-4-14}{2\cdot 3}=\frac{-18}{6}=-3\in \left(-\infty ,-2\right)$  

$x_2=\frac{-4+14}{2\cdot 3}=\frac{10}{6}=\frac{5}{3}\notin \left(-\infty ,-2\right)$  

$II.x\in \left(-2,-1\right)$  

$\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{3}{4}=\frac{2}{-x-1}$  

$\frac{1}{x-1}+\frac{3}{4}=-\frac{2}{x+1}$  

$\frac{4}{4\left(x-1\right)}+\frac{3\left(x-1\right)}{4\left(x-1\right)}=-\frac{2}{x+1}$  

$\frac{4+3\left(x-1\right)}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\frac{4+3x-3}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\frac{3x+1}{4\left(x-1\right)}=-\frac{2}{x+1}$

$\left(3x+1\right)\left(x+1\right)=4\left(x-1\right)\cdot \left(-2\right)$  

$3x^2+3x+x+1=-8x+8$  

$3x^2+12x-7=0$  

$\Delta ={12}^2-4\cdot 3\cdot \left(-7\right)=144+84=228,\sqrt{\Delta }=2\sqrt{57}$  

$x_1=\frac{-12-2\sqrt{57}}{2\cdot 3}=\frac{-6-\sqrt{57}}{3}\notin \left(-2,-1\right)$  

$x_2=\frac{-12+2\sqrt{57}}{2\cdot 3}=\frac{-6+\sqrt{57}}{3}\notin \left(-2,-1\right)$  

$III.x\in \left(-1,+\infty \right)$  

$\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{3}{4}=\frac{2}{x+1}$  

$\frac{1}{x-1}+\frac{3}{4}=\frac{2}{x+1}$  

$\frac{4}{4\left(x-1\right)}+\frac{3\left(x-1\right)}{4\left(x-1\right)}=\frac{2}{x+1}$  

$\frac{4+3\left(x-1\right)}{4\left(x-1\right)}=\frac{2}{x+1}$

$\frac{4+3x-3}{4\left(x-1\right)}=\frac{2}{x+1}$

$\frac{3x+1}{4\left(x-1\right)}=\frac{2}{x+1}$

$\left(3x+1\right)\left(x+1\right)=4\left(x-1\right)\cdot 2$  

$3x^2+3x+x+1=8x-8$  

$3x^2-4x+9=0$  

$\Delta ={\left(-4\right)}^2-4\cdot 3\cdot 9=16-108<0$  

 

Jedyne rozwiązanie tego równania to  x=-3.