$\text{a)}$  

Zał:

$1-x^2\ne 0,1+x\ne 0$  

$x\ne 1,x\ne -1$  

 

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

$1+1-x=2\left(1-x^2\right)$  

$2-x=2-2x^2\mid -2$  

$-x=-2x^2\mid +2x^2$  

$2x^2-x=0$  

$x\left(2x-1\right)=0$  

$x=0\text{lub}2x-1=0$  

$x=0\text{lub}2x=1$  

$x=0\text{lub}x=\frac{1}{2}$  

---

$\text{b)}$     

Zał:

$4x^2+4x+1\ne 0,2x+1\ne 0$  

${\left(2x+1\right)}^2\ne 0,2x+1\ne 0$  

$2x+1\ne 0$  

$2x\ne -1$  

$x\ne -\frac{1}{2}$  

 

$\frac{3-x}{4x^2+4x+1}-\frac{5x}{2x+1}=0\mid \cdot {\left(2x+1\right)}^2$  

$3-x-5x\left(2x+1\right)=0$  

$3-x-10x^2-5x=0$  

$-10x^2-6x+3=0$  

$\Delta ={\left(-6\right)}^2-4\cdot \left(-10\right)\cdot 3=36+120=156$  

$\sqrt{\Delta }=\sqrt{156}=2\sqrt{39}$  

$x_1=\frac{6-2\sqrt{39}}{-20}=\frac{\sqrt{39}-3}{10}$  

$x_2=\frac{6+2\sqrt{39}}{-20}=\frac{-\sqrt{39}-3}{10}$  

---

$\text{c)}$  

Zał:

$x^2-1\ne 0,x^2+2x+1\ne 0$  

$\left(x-1\right)\left(x+1\right)\ne 0,{\left(x+1\right)}^2\ne 0$  

$x\ne 1,x\ne -1$  

 

$\frac{2x}{x^2-1}=\frac{1-x}{x^2+2x+1}$  

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

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

$2x^2+2x=x-1-x^2+x$  

$2x^2+2x=-1-x^2+2x\mid -2x$  

$2x^2=-1-x^2\mid +x^2$  

$3x^2=-1$  

Kwadrat liczby rzeczywistej jest liczbą nieujemną, więc powyższe równanie nie ma rozwiązań w zbiorze liczb rzeczywistych.

---

$\text{d)}$  

Zał:

$x^2+6x+9\ne 0,9-x^2\ne 0$  

${\left(x+3\right)}^2\ne 0,\left(3-x\right)\left(3+x\right)\ne 0$  

$x\ne 3,x\ne -3$  

 

$\frac{2x+1}{x^2+6x+9}+\frac{x-1}{9-x^2}=0$  

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

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

$2x^2-6x+x-3-\left(x^2+3x-x-3\right)=0$  

$2x^2-5x-3-x^2-3x+x+3=0$  

$x^2-7x=0$  

$x\left(x-7\right)=0$  

$x=0\text{lub}x-7=0$  

$x=0\text{lub}x=7$  

---

$\text{e)}$  

Zał:

$x+4\ne 0,x-4\ne 0,x^2-16\ne 0$  

$x+4\ne 0,x-4\ne 0,\left(x-4\right)\left(x+4\right)\ne 0$  

$x\ne -4,x\ne 4$  

 

$\frac{2}{x+4}+\frac{3}{x-4}=\frac{14}{x^2-16}\mid \cdot \left(x-4\right)\left(x+4\right)$  

$2\left(x-4\right)+3\left(x+4\right)=14$  

$2x-8+3x+12=14$  

$5x+4=14$  

$5x=10$  

$x=2$  

---

$\text{f)}$  

Zał:

$x+2\ne 0,x^2-4\ne 0,2x\ne 0$  

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

$x\ne -2,x\ne 2,x\ne 0$  

 

$\frac{3}{x+2}+\frac{12}{x^2-4}+\frac{1}{2x}=0\mid \cdot 2x\left(x-2\right)\left(x+2\right)$  

$3\cdot 2x\left(x-2\right)+12\cdot 2x+\left(x-2\right)\left(x+2\right)=0$  

$6x^2-12x+24x+x^2-4=0$  

$7x^2+12x-4=0$  

$\Delta ={12}^2-4\cdot 7\cdot \left(-4\right)=144+112=256$  

$x_1=\frac{-12-16}{2\cdot 7}=\frac{-28}{14}=-2\notin D$  

$x_2=\frac{-12+16}{2\cdot 7}=\frac{4}{14}=\frac{2}{7}$  

---

$\text{g)}$  

Zał:

$x-4\ne 0,x\ne 0,x^2-4x\ne 0$  

$x\ne 4,x\ne 0,x\left(x-4\right)\ne 0$  

$x\ne 4,x\ne 0$  

 

$\frac{x+5}{x-4}+\frac{3}{x}=\frac{36}{x^2-4x}\mid \cdot x\left(x-4\right)$  

$x\left(x+5\right)+3\left(x-4\right)=36$  

$x^2+5x+3x-12=36$  

$x^2+8x-48=0$  

$\Delta =8^2-4\cdot 1\cdot \left(-48\right)=64+192=256$  

$x_1=\frac{-8-16}{2}=\frac{-24}{2}=-12$  

$x_2=\frac{-8+16}{2}=\frac{8}{2}=4\notin D$  

---

$\text{h)}$  

Zał:

$x+2\ne 0,2-x\ne 0,x^2-4\ne 0$  

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

$x\ne -2,x\ne 2$  

 

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

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

$x^2-2x+x+2=4$  

$x^2-x-2=0$  

$\Delta ={\left(-1\right)}^2-4\cdot 1\cdot \left(-2\right)=1+8=9$  

$x_1=\frac{1-3}{2}=\frac{-2}{2}=-1$  

$x_2=\frac{1+3}{2}=\frac{4}{2}=2\notin D$