Do wyznaczenia dziedziny wystarczy pamiętać, że długości boków muszą być liczbami dodatnimi:

$a)$

$\{\begin{array}{l}x-1>0\\ x+1>0\\ x>0\end{array}\Rightarrow \{\begin{array}{l}x>1\\ x>-1\\ x>0\end{array}\Rightarrow D=\left(1,+\infty \right)$

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

$b)$

$\{\begin{array}{l}x+1>0\\ x+2>0\\ x+3>0\end{array}\Rightarrow \{\begin{array}{l}x>-1\\ x>-2\\ x>-3\end{array}\Rightarrow D=\left(-1,+\infty \right)$

$V\left(x\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x\left(x+2\right)+1\left(x+2\right)\right)\left(x+3\right)=\left(x^2+2x+x+2\right)\left(x+3\right)=$

$=x\left(x^2+3x+2\right)+3\left(x^2+3x+2\right)=x^3+3x^2+2x+3x^2+9x+6=$

$=x^3+6x^2+11x+6$

$c)$

$\{\begin{array}{l}2x+1>0\\ \frac{1}{2}x+1>0\\ 2x-1>0\end{array}\Rightarrow \{\begin{array}{l}x>-\frac{1}{2}\\ x>-2\\ x>\frac{1}{2}\end{array}\Rightarrow D=\left(\frac{1}{2},+\infty \right)$

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

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

$d)$

$\{\begin{array}{l}x+3>0\\ x^2-9>0\end{array}\Rightarrow \{\begin{array}{l}x>-3\\ x^2>9\end{array}\Rightarrow \{\begin{array}{l}x>-3\\ x<-3\vee x>3\end{array}\Rightarrow D=\left(3,+\infty \right)$

$V\left(x\right)=\left(x+3\right)\left(x+3\right)\left(x^2-9\right)={\left(x+3\right)}^2\left(x^2-9\right)=\left(x^2+6x+9\right)\left(x^2-9\right)=$

$=x^2\left(x^2+6x+9\right)-9\left(x^2+6x+9\right)=x^4+6x^3+9x^2-9x^2-54x-81=$

$=x^4+6x^3-54x-81$