Ciao,
1.
$\left [ \left (\frac{3x^{2}-2}{x-1}+\frac{6x-2}{x-3}\times \frac{9-x^{2}}{3x-1} \right ) \times \left ( \frac{1}{x-2}+1 \right )\right ]^{2}\times \left ( 1+\frac{x-2}{x-1} \right )^{-3}$
Determiniamo le condizioni di esistenza, ponendo tutti i denominatori diversi da zero:
$x-1 \neq 0 \rightarrow x \neq 1 $
$x-3 \neq 0 \rightarrow x \neq 3 $
$3x-1 \neq 0 \rightarrow x \neq \frac{1}{3}$
$x-2 \neq 0 \rightarrow x \neq 2 $
C.E.:$ \left \{ x \neq 1, x \neq 3, x \neq \frac{1}{3}, x \neq 2 \right \}$
Prima di iniziare a risolvere l'espressione riscriviamo alcuni polinomi:
$6x-2=2(3x-1)$
$9-x^{2}=(3-x)(3+x)=-(x-3)(x+3)$
Risolviamo l'espressione:
$\left [ \left (\frac{3x^{2}-2}{x-1}+\frac{2(3x-1)}{x-3}\times \frac{-(x-3)(x+3)}{3x-1} \right ) \times \left ( \frac{1}{x-2}+1 \right )\right ]^{2}\times \left ( 1+\frac{x-2}{x-1} \right )^{-3}=$
$\left [ \left (\frac{3x^{2}-2}{x-1}-2(x+3) \right ) \times \left ( \frac{1}{x-2}+1 \right )\right ]^{2}\times \left ( 1+\frac{x-2}{x-1} \right )^{-3}=$
$\left [ \left (\frac{3x^{2}-2-2(x+3)(x-1)}{x-1} \right ) \times \left ( \frac{1+(x-2)}{x-2} \right )\right ]^{2}\times \left ( \frac{(x-1)+x-2}{x-1} \right )^{-3}=$
$\left [ \left (\frac{3x^{2}-2-2(x^{2}-x+3x-3)}{x-1} \right ) \times \left ( \frac{1+x-2}{x-2} \right )\right ]^{2}\times \left ( \frac{x-1+x-2}{x-1} \right )^{-3}=$
$\left [ \left (\frac{3x^{2}-2-2x^{2}+2x-6x+3}{x-1} \right ) \times \left ( \frac{1+x-2}{x-2} \right )\right ]^{2}\times \left ( \frac{x-1+x-2}{x-1} \right )^{-3}=$
$\left [ \left (\frac{x^{2}-4x+4}{x-1} \right ) \times \left ( \frac{x-1}{x-2} \right )\right ]^{2}\times \left ( \frac{2x-3}{x-1} \right )^{-3}=$
$\left [ \left (\frac{(x-2)^{2}}{x-1} \right ) \times \left ( \frac{x-1}{x-2} \right )\right ]^{2}\times \left ( \frac{2x-3}{x-1} \right )^{-3}=$
$\left [ x-2\right ]^{2}\times \left ( \frac{2x-3}{x-1} \right )^{-3}=$
$\left [ x-2\right ]^{2}\times \left ( \frac{x-1}{2x-3} \right )^{3}=$
$\frac{\left ( x-2 \right )^{2}\left ( x-1 \right )^{3}}{\left ( 2x-3 \right )^{3}}$
2.
$\left ( \frac{1}{a-3}+\frac{1}{2a^{2}-3a-9} \right )\times \frac{2a+3}{a^{2}+2a}+\left ( \frac{1}{a^{2}-9}-\frac{1}{a^{2}-3a}\right ) :\frac{1}{a+3} $
Determiniamo le condizioni di esistenza, ponendo tutti i denominatori diversi da zero:
$a-3 \neq 0 \rightarrow a \neq 3 $
$ 2a^{2}-3a-9\neq 0 \rightarrow (2a+3)(a-3)x \neq 0\rightarrow a\neq -\frac{3}{2} , a\neq 3 $
$ a^{2}+2a\neq 0\rightarrow a(a+2)\neq 0\rightarrow a\neq 0,a\neq -2$
$a^{2}-9\neq 0\rightarrow (a-3)(a+3)\neq 0\rightarrow a\neq 3,a\neq -3$
$a^{2}-3a\neq 0\rightarrow a(a-3)\neq 0\rightarrow a\neq 0,a\neq 3$
$a+3 \neq 0 \rightarrow a \neq -3 $
C.E.:$ \left \{ a \neq 0, a \neq \pm3, a \neq -\frac{3}{2}, a \neq- 2 \right \}$
Risolviamo l'espressione:
$\left ( \frac{1}{a-3}+\frac{1}{(2a+3)(a-3)} \right )\times \frac{2a+3}{a(a+2)}+\left ( \frac{1}{(a-3)(a+3)}-\frac{1}{a(a-3)} \right ) :\frac{1}{a+3} =$
$\left ( \frac{2a+3+1}{(2a+3)(a-3)} \right )\times \frac{2a+3}{a(a+2)}+\left ( \frac{a-(a+3)}{a(a-3)(a+3)} \right ) :\frac{1}{a+3} =$
$\left ( \frac{2a+4}{(2a+3)(a-3)} \right )\times \frac{2a+3}{a(a+2)}+\left ( \frac{a-a-3}{a(a-3)(a+3)} \right ) :\frac{1}{a+3} =$
$\left ( \frac{2(a+2)}{(2a+3)(a-3)} \right )\times \frac{2a+3}{a(a+2)}+\left ( \frac{-3}{a(a-3)(a+3)} \right ) :\frac{1}{a+3} =$
$\frac{2}{a(a-3)} +\left ( \frac{-3}{a(a-3)(a+3)} \right ) \times \left (a+3 \right ) =$
$\frac{2}{a(a-3)} -\frac{3}{a(a-3)}=$
$\frac{2-3}{a(a-3)} =$
$- \frac{1}{a(a-3)}$
saluti 🙂