a) Założenia:

$x^2+x-2\ne 0\wedge x+1\ne 0$  

$x^2-x+2x-2\ne 0\wedge x+1\ne 0$  

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

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

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

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

 

$\text{Dla}x\in \left(-\infty ,-2\right)$  

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

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

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

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

$\frac{x-3}{x^2-1}=-\frac{3}{4}$  

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

$4x-12=-3x^2+3$  

$3x^2+4x-15=0$  

$\Delta =4^2-4\cdot 3\cdot \left(-15\right)=16+180=196$  

$\sqrt{\Delta }=14$  

$x_1=\frac{-4-14}{6}\vee x_2=\frac{-4+14}{6}$  

$x_1=-3\vee x_2=\frac{10}{6}$  

Jedynym rozwiązaniem w badanym przedziale jest liczba -3.

 

$\text{Dla}x\in \left(-2,-1\right)$  

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

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

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

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

$\frac{3x-1}{x^2-1}=-\frac{3}{4}$   

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

$12x-4=-3x^2+3$   

$3x^2+12x-7=0$    

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

$\sqrt{228}=2\sqrt{57}$  

$x_1=\frac{-12-2\sqrt{57}}{6}=\frac{-6-\sqrt{57}}{3}\approx -4,5$    

 

$x_2=\frac{-12+2\sqrt{57}}{6}=\frac{-6+\sqrt{57}}{3}\approx 0,5$      

brak rozwiązań w tym przedziale.

 

$\text{Dla}x\in \left(-1,\infty \right)\backslash \left\{1\right\}$  

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

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

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

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

$\frac{3-x}{x^2-1}=-\frac{3}{4}$  

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

$12-4x=-3x^2+3$  

$3x^2-4x+9=0$  

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

brak rozwiązań. 

 

Jedynym rozwiązaniem jest liczba  $-3$