If \( (2 a-5)(3 a-2)=6 a^{2}+k a+10 \), then \( k= \) \( \begin{array}{llll}\text { (a) } 15 & \text { (b) } 19 & \text { (c) }-19 & \text { (d) } 6\end{array} \)

For the following system to be consistent, \( \left\{\begin{array}{l}6 x_{1}-3 x_{2}-2 x_{3}=-6 \\ 5 x_{1}-8 x_{2}+k x_{3}=-4 \\ -7 x_{1}+4 x_{2}+7 x_{3}=8\end{array}\right. \) we must have \( k \neq \square \) You may enter your answer as a fraction.

Simplificar y obtener la minima expresión: \( \frac{\frac{b}{a}+\frac{a-2 b}{b}}{\left(\frac{a^{2}+3 a b+3 b^{2}}{b}+\frac{b^{2}}{a}\right) \cdot(a-b)^{2}} \)

b) \( 5^{0}+\frac{1}{3} \cdot\left[\left(3-\frac{1}{5}\right) \cdot 5^{-2}+2 \cdot\left(1-\frac{2}{5}\right)^{3}\right]: \frac{4}{3} \)

42. \( \frac{3 x+y}{2 x+3+z} \quad \operatorname{per} x=-\frac{1}{9} \quad y=\frac{1}{2} \quad, \quad z=\frac{2}{9} \)

\begin{tabular}{l} B) DIRECTIONS: Solve each equation using the QUADRATIC FORMULA: \( x= \) \\ \( \qquad \)\begin{tabular}{|l|l|l}\hline 5\( ] c^{2}-5 c+4=0 \) & \( 6] b^{2}-b-12=0 \) & \( 73 a^{2}-4 a-12=0 \) \\ \hline\end{tabular} \\ \hline\end{tabular}