Question

Numarul termenilor rationali ai dezvoltarii​

Answer 1

$\it T_{k+1}=C^k _{30}(2^{\frac{1}{2}})^{30-k}\cdot(5^{\frac{1}{3}})^k\\ \\ 2^{\dfrac{30-k}{2}}\in\mathbb{Q} \Rightarrow k \in M_2\ \ \ \ (1)\\ \\ \\ 5^{\dfrac{k}{3}} \Rightarrow k\in M_3\ \ \ \ (2)\\ \\ \\ k\leq30\ \ \ \ (3)\\ \\ \\ (1),\ (2),\ (3)\Rightarrow k\in \{0,\ 6,\ 12,\ 18,\ 24,\ 30 \}$

Deci,  dezvoltarea conține 6 termeni raționali.

Answer 2

Răspuns:

6 termeni raționali.

Explicație pas cu pas:

$(\sqrt 2 + \sqrt[3]{5})^{30}\\ \\ T_{k+1} = C_{30}^{k}\cdot \sqrt{2}^{30-k}\cdot \sqrt[3]{5}^{k} \\ \\ =C_{30}^k\cdot 2^{\frac{30-k}{2}}\cdot 5^{\frac{k}{3}}\\ \\ \\ \Rightarrow 30-k= M_{2}\quad \text{si}\quad k = M_3\\\\\Rightarrow k = 30 - M_{2} \quad \text{si}\quad k = M_3 \\ \\ \Rightarrow k = 30 - 2a,\,\,\, a\in \mathbb{N}\quad \text{si}\quad k = M_3\\ \\ \Rightarrow 30-2a = M_3 \Rightarrow 2a = 30-M_3 \Rightarrow$

$\Rightarrow a = \dfrac{30-M_3}{2} \Rightarrow a \in\Big\{\dfrac{30}{2},\dfrac{27}{2},\dfrac{24}{2},\dfrac{21}{2},\dfrac{18}{2},...,\dfrac{3}{2},\dfrac{0}{2}\Big\}\\ \\\\ \text{In multimea }a\text{ de la }30,24,18,12,6,0\,\Rightarrow \boxed{6}\text{ numere naturale.}$