a)

Zał:

$x-1>0\wedge x-1\ne 1\wedge 3x-1>0$  

$x>1\wedge x\ne 2\wedge x>\frac{1}{3}$  

$x\in \left(1,2\right)\cup \left(2,+\infty \right)$  

 

${\log }_{x-1}\left(3x-1\right)=3$     

${\left(x-1\right)}^3=3x-1$      

$x^3-3x^2+3x-1=3x-1\mid -3x+1$  

$x^3-3x^2=0$  

$x^2\left(x-3\right)=0$  

$x^2=0\vee x-3=0$  

$x=0\notin D\vee x=3$  

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b)

Zał:

$x-1>0\wedge x-1\ne 1\wedge 2x^2+4x-6>0$  

$x>1\wedge x\ne 2\wedge x^2+2x-3>0$  

$x>1\wedge x\ne 2\wedge \Delta =2^2-4\cdot 1\cdot \left(-3\right)=4+12=16$  

$x>1\wedge x\ne 2\wedge x_1=\frac{-2-4}{2}=\frac{-6}{2}=-3,x_2=\frac{-2+4}{2}=\frac{2}{2}=1$  

$x>1\wedge x\ne 2\wedge \left(x+3\right)\left(x-1\right)>0$  

$x>1\wedge x\ne 2\wedge x\in \left(-\infty ,-3\right)\cup \left(1,+\infty \right)$  

$x\in \left(1,2\right)\cup \left(2,+\infty \right)$  

 

${\log }_{x-1}\left(2x^2+4x-6\right)=2$  

${\left(x-1\right)}^2=2x^2+4x-6$  

$x^2-2x+1=2x^2+4x-6\mid -x^2+2x-1$  

$0=x^2+6x-7$  

$\Delta =6^2-4\cdot 1\cdot \left(-7\right)=36+28=64$  

$x_1=\frac{-6-8}{2}=\frac{-14}{2}=-7\notin D$  

$x_2=\frac{-6+8}{2}=\frac{2}{2}=1\notin D$  

Zatem jest to równanie sprzeczne.

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c)

Zał:

$x>0\wedge x\ne 1\wedge 4x^2+x-4>0$  

$x>0\wedge x\ne 1\wedge \Delta =1^2-4\cdot 4\cdot \left(-4\right)=1+64=65$  

$x>0\wedge x\ne 1\wedge x_1=\frac{-1-\sqrt{65}}{8},x_2=\frac{-1+\sqrt{65}}{8}$  

$x>0\wedge x\ne 1\wedge x\in \left(-\infty ,\frac{-1-\sqrt{65}}{8}\right)\cup \left(\frac{-1+\sqrt{65}}{8},+\infty \right)$  

$x\in \left(\frac{-1+\sqrt{65}}{8},1\right)\cup \left(1,+\infty \right)$  

 

${\log }_x\left(4x^2+x-4\right)=3$  

$x^3=4x^2+x-4\mid -4x^2-x+4$  

$x^3-4x^2-x+4=0$  

$x^2\left(x-4\right)-1\left(x-4\right)=0$  

$\left(x^2-1\right)\left(x-4\right)=0$  

$\left(x-1\right)\left(x+1\right)\left(x-4\right)=0$  

$x-1=0\vee x+1=0\vee x-4=0$  

$x=1\notin D\vee x=-1\notin D\vee x=4$  

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d)

Zał:

$x>0\wedge x\ne 1\wedge \frac{x+2}{x}>0$  

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

$x>0\wedge x\ne 1\wedge x\in \left(-\infty ,-2\right)\cup \left(0,+\infty \right)$  

$x\in \left(0,1\right)\cup \left(1,+\infty \right)$  

 

${\log }_x\frac{x+2}{x}=1$  

$x=\frac{x+2}{x}\mid \cdot x$  

$x^2=x+2\mid -x-2$  

$x^2-x-2=0$  

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

$x_1=\frac{1-3}{2}=\frac{-2}{2}=-1\notin D$  

$x_2=\frac{1+3}{2}=\frac{4}{2}=2$  

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e)

Zał:

$x^2>0\wedge x^2\ne 1\wedge x+2>0$  

$x\ne 0\wedge x\ne 1,x\ne -1\wedge x>-2$  

$x\in \left(-2,-1\right)\cup \left(-1,0\right)\cup \left(0,1\right)\cup \left(1,+\infty \right)$  

 

${\log }_{x^2}\left(x+2\right)=1$  

$x^2=x+2\mid -x-2$  

$x^2-x-2=0$  

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

$x_1=\frac{1-3}{2}=\frac{-2}{2}=-1\notin D$  

$x_2=\frac{1+3}{2}=\frac{4}{2}=2$  

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f)

Zał:

$4-x^2>0\wedge 4-x^2\ne 1$  

$-\left(x-2\right)\left(x+2\right)>0\wedge 3\ne x^2$  

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

$x\in \left(-2,2\right)\wedge x\ne \sqrt{3},x\ne -\sqrt{3}$  

$x\in \left(-2,-\sqrt{3}\right)\cup \left(-\sqrt{3},\sqrt{3}\right)\cup \left(\sqrt{3},2\right)$  

 

${\log }_{4-x^2}16=2$  

${\left(4-x^2\right)}^2=16$  

$4-x^2=4$  

$x^2=0$  

$x=0$