Варнант 3 1. Выполните действия \( \begin{array}{ll}\frac{2 x+8}{6 x-12}+\frac{6-2 x}{4 x-8} & \frac{2 x+8}{4 x-12}-\frac{2 x-2}{x-3} \\ \frac{52 x^{9}}{102 y^{7}} \cdot \frac{68 y^{5}}{78 x^{6}} & \frac{64 a^{7}}{45 y^{10}}: \frac{6 a^{5}}{135 y^{6}}\end{array} \)

\( 2 x - \frac { 16 x + 1 } { 5 ^ { x + 1 } } + 1 = 0 \)

2 varies directly as the cube of \( x \) and inversely as \( y \). If \( z=237 \) when \( x=10 \) and \( y=2 \), find \( z \) if \( x=8 \) and Round your answer to two decimal places if necessary.

Step 2 of 2: Find the value of \( f(a) \). \( \quad f(x)=4 x^{3}+4 x \)

\( \left. \begin{array} { l } { - x + y = 3 } \\ { y = 3 - x } \\ { y = x - 3 } \end{array} \right. \)

tinúa con el ejercicio de cramer \[ x=\frac{\left|\begin{array}{ll}5 & 3 \\ 27 & 7\end{array}\right|}{\left|\begin{array}{ll}5 & 3 \\ 4 & 7\end{array}\right|}=\frac{(5)(7)-(27)(3)}{(5)(7)-(4)(3)} \Rightarrow \frac{35-81}{35-12}= \] \( \begin{array}{l}a_{1} x+b_{1} y=c_{1} \\ a_{2} x+b_{2} y=c_{2}\end{array} \) \( \begin{array}{l}x+3 y=5 \\ x+7 y=27\end{array} \)