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Одредити област дефинисаности функција

пр.1)  $y =  - 3{x^4} + 2{x^3} - 7{x^2} + 5x - 5$

пр.2)  $y = \frac{{ - 2{x^2} + 1}}{{{x^2} - 1}}$

пр.3)  $y = \sqrt {{x^2} - 7x + 10} $

пр.4)  $y = \sqrt[3]{{ - 5{x^5}+2{x^3}+1}}$

пр.5)  $y = log(2{x^2}-x-10)$

пр.1)  $y =  - 3{x^4} + 2{x^3} - 7{x^2} + 5x - 5$

\[Df:\mathbb{R}\]

пр.2)  $y = \frac{{ - 2{x^2} + 1}}{{{x^2} - 1}}$

\[\begin{gathered}
Df:{x^2} - 1 \ne 0 \hfill \\
{x^2} \ne 1 \hfill \\
x \ne \pm 1 \hfill \\
Df:\mathbb{R}\backslash \left\{ { - 1,1} \right\} \hfill \\
\end{gathered} \]

пр.3)  $y = \sqrt {{x^2} - 7x + 10} $

\[\begin{gathered}
Df:{x^2} - 7x + 10 \geqslant 0 \hfill \\
{x^2} - 7x + 10 = 0 \hfill \\
{x_{1,2}} = \frac{{7 \pm \sqrt {49 - 40} }}{2} \hfill \\
{x_{1,2}} = \frac{{7 \pm 3}}{2} \hfill \\
{x_1} = 2,{x_2} = 5 \hfill \\
\end{gathered} \]

\[Df:x \in \left( { - \infty ,2} \right] \cup \left[ {5, + \infty } \right)\]

пр.4)  $y = \sqrt[3]{{ - 5{x^5}+2{x^3}+1}}$

\[Df:\mathbb{R}\]

пр.5)  $y = log(2{x^2}-x-10)$

\[\begin{gathered}
Df:2{x^2} - x - 10 > 0 \hfill \\
2{x^2} - x - 10 = 0 \hfill \\
{x_{1,2}} = \frac{{1 \pm \sqrt {1 + 80} }}{4} = \frac{{1 \pm 9}}{4} \hfill \\
{x_1} = - 2,{x_1} = \frac{5}{2} \hfill \\
\end{gathered} \]

\[Df:x \in \left( { - \infty , - 2} \right] \cup \left[ {\frac{5}{2}, + \infty } \right)\]