a) Zał:

$\frac{x}{2-x}>0\wedge 2-x\ne 0$  

$x\left(2-x\right)>0\wedge x\ne 2$  

$-x\left(x-2\right)>0\mid :\left(-1\right)\wedge x\ne 2$  

$x\left(x-2\right)<0\wedge x\ne 2$  

$x\in \left(0,2\right)$  

 

${\log }_{0,25}\frac{x}{2-x}>-1$  

${\log }_{0,25}\frac{x}{2-x}>{\log }_{0,25}4$  

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

$x\left(2-x\right)<4{\left(2-x\right)}^2\mid -4{\left(2-x\right)}^2$  

$x\left(2-x\right)-4{\left(2-x\right)}^2<0$  

$\left(2-x\right)\cdot \left(x-4\left(2-x\right)\right)<0$  

$\left(2-x\right)\left(x-8+4x\right)<0$  

$-\left(x-2\right)\left(5x-8\right)<0\mid :\left(-1\right)$  

$5\left(x-2\right)\left(x-\frac{8}{5}\right)>0$  

$x\in \left(-\infty ,\frac{8}{5}\right)\cup \left(2,+\infty \right)$  

Uwzględniając założenie otrzymujemy:

$x\in \left(0,1\frac{3}{5}\right)$  

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b) Zał:

$\frac{x+1}{x-1}>0\wedge x-1\ne 0$  

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

$x\in \left(-\infty ,-1\right)\cup \left(1,+\infty \right)$  

 

${\log }_{\frac{1}{5}}\frac{x+1}{x-1}<0$  

${\log }_{\frac{1}{5}}\frac{x+1}{x-1}<{\log }_{\frac{1}{5}}1$  

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

$\left(x+1\right)\left(x-1\right)>{\left(x-1\right)}^2\mid -{\left(x-1\right)}^2$  

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

$\left(x-1\right)\cdot \left(\left(x+1\right)-\left(x-1\right)\right)>0$  

$\left(x-1\right)\left(x+1-x+1\right)>0$  

$2\left(x-1\right)>0$  

$x>1$  

$x\in \left(1,+\infty \right)$  

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c) Zał:

$\frac{x+2}{1-x}>0\wedge 1-x\ne 0$  

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

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

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

$x\in \left(-2,1\right)$  

 

${\log }_8\frac{x+2}{1-x}\ge 1$  

${\log }_8\frac{x+2}{1-x}\ge {\log }_{8}8$  

$\frac{x+2}{1-x}\ge 8\mid \cdot {\left(1-x\right)}^2$  

$\left(x+2\right)\left(1-x\right)\ge 8{\left(1-x\right)}^2\mid -8{\left(1-x\right)}^2$  

$\left(x+2\right)\left(1-x\right)-8{\left(1-x\right)}^2\ge 0$  

$\left(1-x\right)\cdot \left(\left(x+2\right)-8\left(1-x\right)\right)\ge 0$  

$\left(1-x\right)\left(x+2-8+8x\right)\ge 0$  

$-\left(x-1\right)\left(9x-6\right)\ge 0\mid :\left(-1\right)$  

$9\left(x-1\right)\left(x-\frac{2}{3}\right)\le 0$  

$x\in ⟨\frac{2}{3},1⟩$  

Uwzględniając założenie otrzymujemy:

$x\in ⟨\frac{2}{3},1)$