$$ \begin{flalign*} &(a) \quad \text{Berechne: } \int_{0}^{1} \frac{x^ §§V2(3,15,3)§§ - 1}{\ln x} \, \mathrm{d}x && \\ &(b) \quad \text{Löse die Gleichung: } \sqrt{ §§V2(3,15,3)§§ x-1}+\sqrt[ §§V2(3,15,3)§§ ]{ §§V2(2,10,2)§§ x-5}= §§V2(3,15,3)§§ && \\ &(c) \quad \text{Berechne: } \frac{ §§V2(2,10,2)§§ }{5} \times \left(\frac{ §§V2(3,15,3)§§ }{4} + \frac{5}{6}\right) && \\ &(d) \quad \text{Löse die Ungleichung: } §§V2(2,10,2)§§ x + 5 \geq §§V2(3,15,3)§§ x - §§V2(2,10,2)§§ && \\ &(e) \quad \text{Berechne: } \left(\frac{ §§V2(3,15,3)§§ }{4}\right)^ §§V2(2,10,2)§§ \cdot \left(\frac{ §§V2(2,10,2)§§ }{ §§V2(3,15,3)§§ }\right)^ §§V2(3,15,3)§§ && \\ &(f) \quad \text{Löse die Gleichung: } \frac{ §§V2(2,10,2)§§ x}{ §§V2(3,15,3)§§ } - \frac{5}{ §§V2(2,10,2)§§ } = \frac{x+4}{6} && \\ &(g) \quad \text{Berechne: } \sqrt{16} \cdot \left(\frac{ §§V2(2,10,2)§§ }{ §§V2(3,15,3)§§ }\right)^{- §§V2(2,10,2)§§ } && \\ &(h) \quad \text{Löse das Gleichungssystem:} \\ &\quad\quad \begin{cases} §§V2(2,10,2)§§ x - §§V2(3,15,3)§§ y = 5 \\ 4x + y = §§V2(2,10,2)§§ \end{cases} && \\ &(i) \quad \text{Berechne: } \frac{5}{8} - \left(\frac{1}{ §§V2(3,15,3)§§ } - \frac{ §§V2(2,10,2)§§ }{5}\right) && \\ &(j) \quad \text{Löse die Gleichung: } §§V2(2,10,2)§§ ( §§V2(3,15,3)§§ x - 1) = 4x + 5 &&\\ &(a12) \quad \text{Berechne: } \frac{ §§V2(3,15,3)§§ }{4} \cdot \left(\frac{5}{6} + \frac{7}{8}\right) && \\ &(b12) \quad \text{Löse die Gleichung: } \frac{ §§V2(2,10,2)§§ x- §§V2(3,15,3)§§ }{4} - \frac{ §§V2(3,15,3)§§ x+2}{ §§V2(2,10,2)§§ } = \frac{x+1}{ §§V2(3,15,3)§§ } && \\ &(c12) \quad \text{Berechne: } \sqrt[ §§V2(3,15,3)§§ ]{\frac{ §§V2(2,10,2)§§ 7}{64}} && \\ &(d12) \quad \text{Löse das Gleichungssystem:} \\ &\quad\quad \begin{cases} §§V2(2,10,2)§§ x + §§V2(3,15,3)§§ y = §§V1(1,10,1)§§ 0 \\ 4x - y = 5 \end{cases} && \\ &(e §§V1(1,10,1)§§ §§V2(2,10,2)§§ ) \quad \text{Berechne: } \left(\frac{5}{6}\right)^{- §§V2(2,10,2)§§ } \cdot \left(\frac{7}{10}\right)^{- §§V1(1,10,1)§§ } && \\ &(f12) \quad \text{Löse die Ungleichung: } \frac{ §§V2(2,10,2)§§ x- §§V1(1,10,1)§§ }{ §§V2(3,15,3)§§ } > \frac{x+ §§V2(2,10,2)§§ }{4} && \\ &(g12) \quad \text{Berechne: } \log_{ §§V2(2,10,2)§§ } 8 + \log_{\frac{ §§V1(1,10,1)§§ }{ §§V2(2,10,2)§§ }} 16 && \\ &(h12) \quad \text{Löse die Gleichung: } \sqrt{ §§V2(2,10,2)§§ x+ §§V1(1,10,1)§§ } = \sqrt{ §§V2(3,15,3)§§ x- §§V2(2,10,2)§§ } && \\ &(i12) \quad \text{Berechne: } \sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{6}\right) + \tan\left(\frac{\pi}{ §§V2(3,15,3)§§ }\right) && \\ &(j12) \quad \text{Löse das Gleichungssystem:} \\ &a1) \quad \text{Berechne: } \frac{ §§V2(3,15,3)§§ }{4} \cdot \left(\frac{5}{6} + \frac{7}{8}\right) && \\ &a2) \quad \text{Löse die Gleichung: } \frac{ §§V2(2,10,2)§§ x- §§V2(3,15,3)§§ }{4} - \frac{ §§V2(3,15,3)§§ x+ §§V2(2,10,2)§§ }{ §§V2(2,10,2)§§ } = \frac{x+ §§V1(1,10,1)§§ }{ §§V2(3,15,3)§§ } && \\ &a3) \quad \text{Berechne: } \sqrt[3]{\frac{ §§V2(2,10,2)§§ 7}{64}} && \\ &a4) \quad \text{Löse das Gleichungssystem:} \\ &\quad\quad \begin{cases} §§V2(2,10,2)§§ x + §§V2(3,15,3)§§ y = §§V1(1,10,1)§§ 0 \\ 4x - y = 5 \end{cases} && \\ &a5) \quad \text{Berechne: } \left(\frac{5}{6}\right)^{-2} \cdot \left(\frac{7}{10}\right)^{-1} && \\ &a6) \quad \text{Löse die Ungleichung: } \frac{ §§V2(2,10,2)§§ x- §§V1(1,10,1)§§ }{ §§V2(3,15,3)§§ } > \frac{x+ §§V2(2,10,2)§§ }{4} && \\ &a7) \quad \text{Berechne: } \log_{2} 8 + \log_{\frac{ §§V1(1,10,1)§§ }{ §§V2(2,10,2)§§ }} 16 && \\ &a8) \quad \text{Löse die Gleichung: } \sqrt{ §§V2(2,10,2)§§ x+ §§V1(1,10,1)§§ } = \sqrt{ §§V2(3,15,3)§§ x- §§V2(2,10,2)§§ } && \\ &a9) \quad \text{Berechne: } \sin\left(\frac{\pi}{4}\right) \cdot \cos\left(\frac{\pi}{6}\right) + \tan\left(\frac{\pi}{ §§V2(3,15,3)§§ }\right) && \\ &a10) \quad \text{Löse das Gleichungssystem:} \\ &\quad\quad \begin{cases} §§V2(2,10,2)§§ x - y + z = 4 \\ x + §§V2(3,15,3)§§ y - §§V2(2,10,2)§§ z = - §§V1(1,10,1)§§ \\ §§V2(3,15,3)§§ x + y + 4z = 9 \end{cases} && \\ &(b1) \quad \text{Berechne: } \frac{5}{8} \cdot \left(\frac{ §§V2(3,15,3)§§ }{4} + \frac{ §§V2(2,10,2)§§ }{5}\right) && \\ &(b2) \quad \text{Löse die Gleichung: } §§V2(2,10,2)§§ x^2 + 5x - §§V2(3,15,3)§§ = 0 && \\ &(b3) \quad \text{Berechne: } \sqrt{144} && \\ &(b4) \quad \text{Löse das Gleichungssystem:} \\ &\quad\quad \begin{cases} §§V2(3,15,3)§§ x - §§V2(2,10,2)§§ y = 7 \\ 5x + 4y = 11 \end{cases} && \\ &(b5) \quad \text{Berechne: } \left(\frac{ §§V2(3,15,3)§§ }{5}\right)^{-2} \cdot \left(\frac{4}{7}\right)^{- §§V1(1,10,1)§§ } && \\ &(b6) \quad \text{Löse die Ungleichung: } \frac{ §§V2(2,10,2)§§ x+ §§V1(1,10,1)§§ }{ §§V2(3,15,3)§§ } \geq \frac{x- §§V2(2,10,2)§§ }{ §§V2(2,10,2)§§ } && \\ &(b7) \quad \text{Berechne: } \log_{2} §§V2(3,15,3)§§ 2 + \log_{\frac{ §§V1(1,10,1)§§ }{2}} 4 && \\ &(b8) \quad \text{Löse die Gleichung: } \sqrt{4x+ §§V1(1,10,1)§§ } = \sqrt{5x- §§V2(2,10,2)§§ } && \\ &(b9) \quad \text{Berechne: } \sin\left(\frac{\pi}{6}\right) \cdot \cos\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{ §§V2(3,15,3)§§ }\right) && \\ \end{flalign*} \textbf{Ferrarijeva rezolventa} Given the quartic equation: \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \] where $a \neq 0$, the Ferrari's method aims to find the solutions $x$ in terms of the coefficients $a$, $b$, $c$, $d$, and $e$. To solve the quartic equation, follow these steps: \begin{enumerate} \item Calculate the discriminant $D$: \[ D = b^2 - 4ac \] \item Find the coefficients $p$ and $q$: \[ p = \frac{8ac - 3b^2}{8a^2}, \quad q = \frac{b^3 - 4abc + 8a^2d}{8a^3} \] \item Calculate the intermediate value $m$: \[ m = \sqrt[3]{\frac{-q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} \] \item Find the three roots $x_1$, $x_2$, and $x_3$ using the following expressions: \begin{align*} x_1 &= \frac{-b}{4a} + \frac{1}{2\sqrt{a}}\sqrt{-2p + \frac{2}{m} + \frac{3bm}{a}} \\ x_2 &= \frac{-b}{4a} - \frac{1}{2\sqrt{a}}\sqrt{-2p + \frac{2}{m} - \frac{3bm}{a}} \\ x_3 &= \frac{-b}{4a} + \frac{1}{2\sqrt{a}}\sqrt{-2p - \frac{2}{m} + \frac{3bm}{a}} \end{align*} \item If $D \neq 0$, find the fourth root $x_4$ using: \[ x_4 = \frac{-b}{4a} - \frac{1}{2\sqrt{a}}\sqrt{-2p - \frac{2}{m} - \frac{3bm}{a}} \] \end{enumerate}