Question

Si a+∛3=1+3/∛3, entonces el valor de L=a^3-3a^2+12a-16 es

Answer 1

Respuesta:

$\int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \alpha \alpha \sqrt{x} \leq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$$\beta \beta \beta \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] x_{123} \sqrt[n]{x} \frac{x}{y} \sqrt{x}$$x_{123} \alpha \pi \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \beta \left \{ {{y=2} \atop {x=2}} \right.$$\sqrt[n]{x} \sqrt[n]{x} \frac{x}{y} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \geq \geq \frac{x}{y} \frac{x}{y} x^{2} x^{2} \\ x^{2} \beta \int\limits^a_b {x} \, dx \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

Explicación paso a paso:

es 10