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Explicație pas cu pas:

(∛2+ordin 4√4)^100=(2^(1/3)+4^(1/4))^100

T(k+1)=C n luate cate k * 2^(100-k)/3*2^(k/2)

100-k trebuie sa fie multiplu de 4

adica 100-k=multiplu 3 adica k=100-multiplu3

sau 100,97,94,91,,,,,3,0 (1)

si avem k multiplu de 2

adica: 100,98,92,90,88,,,,,2,0 (2)

Din 1 intersectat cu 2 avem ca k trebuie sa fie si multiplu de 3 si de 2 adica de 6

deci am avea:

100,94,88,82,,,4

pentru k=1,2,3 ,0 nu divide 100

n-nr de termeni

n=(100-4)/6+1=96/6+1=16+1=17 termeni rationali

Bafta!

[tex](\sqrt[3]{3}+\sqrt[4]{4})^{100} = (3^{\frac{1}{3}}+2^\frac{1}{2})^{100}\\ \\ T_{k+1}=C_n^k\cdot a^{n-k}\cdot b^{k} = C_n^k \cdot 3^{\frac{100-k}{3}}\cdot 2^{\frac{k}{2}} \\ \\ \boxed{1}\quad 100-k \in \{0,3,6,9,12,15,18,21,24,...\}\Rightarrow \\ \\ \Rightarrow -k \in \{-100,-97,-94,-91,-88,-85,-....-4,-1\} \Rightarrow[/tex]

[tex]\Rightarrow k\in \{1,4,7,10,13,...,94,97,100\},\quad k = 3c-2,\,\, c = \overline{1,2,3,...}\\ \\ \boxed{2}\quad k\in \{0,2,4,6,8,10,...,98,100\} \\ \\ \text{Din (1) si (2) }\Rightarrow 3c-2 \in D_2 \Rightarrow\\ \\ \Rightarrow 3c-2 \in\{0,2,4,6,8,10,12,14,18,...,100\} \Rightarrow \\ \\ \Rightarrow 3c \in \{2,4,6,8,10,12,14,16,18,...,102\} \Rightarrow \\ \\ \Rightarrow c \in \{\_,\_,2\_,\_,4,\_,\_,6,\_,\_,8,...,34\}[/tex]

De la 2 la 34 (din 2 in 2) sunt 17 numere.

2(1,2,3,4,...,17) => 17 numere

=> Sunt 17 termeni rationali.