Hasil dari \( 3\left(x-\frac{1}{x}\right)^{2} \) adalah ... \( \begin{array}{ll}\text { A. } 3 x^{2}-6 x+\frac{3}{x^{2}} & \text { C. } 3 x^{2}-6+\frac{3}{x^{2}} \\ \text { B. } 3 x^{2}-2 x+\frac{1}{x^{2}} & \text { D. } 3 x^{2}-2+\frac{1}{x^{2}}\end{array} \)

7. \( 2 x^{2}+x-4=0 \) \( \begin{array}{ll}\text { a. }-\frac{1}{2} \pm \frac{\sqrt{33}}{4} & \text { c. }-\frac{1}{4} \pm \frac{\sqrt{33}}{4} \\ \text { b. }-4 \pm \frac{\sqrt{66}}{4} & \text { d. }-\frac{1}{2} \pm \frac{\sqrt{33}}{2}\end{array} \)

Solve the inequality. Express the solution set in set-builder notation and interval notation. Graph the solution set on a real number line. \[ 5(x-6)>4(x-6)+x \]

3. (*) Dado los vectores \( \vec{a}=\left(1,2 x_{1}, 3\right) \), \( \vec{b}=\left(-1,3 x_{2},-2 x_{1}\right), \vec{c}=\left(2,2,-3 x_{2}\right) \) los valores de \( x_{1} \quad y x_{2} \) para los cuales se cumple que \( 2 \vec{a}-3 \vec{b}+\vec{c}=(7,15,21) \) es respectivamente: a. \( (2,-1) \) b. \( (-1,2) \) c. \( (2,-3) \) d. \( (-3,1) \)

Graph the parabola. \[ y=(x+2)^{2}+4 \] Plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. Then click on the graph-a-function button.

в) \( x+15 \frac{3}{17}=20 \frac{14}{17} \) \( x= \) \( x= \) \( x= \)