Qual o valor da expressão no anexo?
Como eu posso trabalhar com essa P.A.?
Anexos:
Soluções para a tarefa
Respondido por
0
1+2+3+...+50 é uma PA de razão 1 e tem 50 termos. A soma é:
(1+50)*50/2 = 1275
5+10+15+...+250 é uma PA de razão 5 e tem 50 termos. A soma é:
(5+250)*50/2 = 6375
Então fica:
(1275/6375) = 1/5
![(\dfrac{1}{5})^{- \dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{1,25}})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{ \frac{125}{100} }})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{ \frac{5}{4} }})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{ \frac{2}{2}* \sqrt{5}})^{-1} \\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{ \sqrt{5}})^{-1} \\\\
(5)^{\dfrac{1}{2}}*( \sqrt[6]{5})^{-1} \\\\
(5)^{\dfrac{1}{2}}*({5^\dfrac{1}{6}} )^{-1}\\\\
5^{\dfrac{1}{2}}*{5^\dfrac{-1}{6}} \\\\\\
5^{\dfrac{1}{2}-\dfrac{1}{6}}](https://tex.z-dn.net/?f=+%28%5Cdfrac%7B1%7D%7B5%7D%29%5E%7B-+%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B3%5D%7B2+%5Csqrt%7B1%2C25%7D%7D%29%5E%7B-1%7D%5C%5C%5C%5C%0A+%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B3%5D%7B2+%5Csqrt%7B+%5Cfrac%7B125%7D%7B100%7D+%7D%7D%29%5E%7B-1%7D%5C%5C%5C%5C%0A%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B3%5D%7B2+%5Csqrt%7B+%5Cfrac%7B5%7D%7B4%7D+%7D%7D%29%5E%7B-1%7D%5C%5C%5C%5C%0A%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B3%5D%7B+%5Cfrac%7B2%7D%7B2%7D%2A++%5Csqrt%7B5%7D%7D%29%5E%7B-1%7D+%5C%5C%5C%5C%0A%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B3%5D%7B+%5Csqrt%7B5%7D%7D%29%5E%7B-1%7D+%5C%5C%5C%5C%0A%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28+%5Csqrt%5B6%5D%7B5%7D%29%5E%7B-1%7D+%5C%5C%5C%5C%0A%0A%285%29%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%28%7B5%5E%5Cdfrac%7B1%7D%7B6%7D%7D+%29%5E%7B-1%7D%5C%5C%5C%5C%0A%0A5%5E%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2A%7B5%5E%5Cdfrac%7B-1%7D%7B6%7D%7D+%5C%5C%5C%5C%5C%5C%0A%0A5%5E%7B%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B1%7D%7B6%7D%7D+ "(\dfrac{1}{5})^{- \dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{1,25}})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{ \frac{125}{100} }})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{2 \sqrt{ \frac{5}{4} }})^{-1}\\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{ \frac{2}{2}* \sqrt{5}})^{-1} \\\\
(5)^{\dfrac{1}{2}}*( \sqrt[3]{ \sqrt{5}})^{-1} \\\\
(5)^{\dfrac{1}{2}}*( \sqrt[6]{5})^{-1} \\\\
(5)^{\dfrac{1}{2}}*({5^\dfrac{1}{6}} )^{-1}\\\\
5^{\dfrac{1}{2}}*{5^\dfrac{-1}{6}} \\\\\\
5^{\dfrac{1}{2}-\dfrac{1}{6}}")
![5^{\dfrac{3-1}{6}}\\\\ 5^{\dfrac{2}{6}}\\\\
5^{\dfrac{1}{3}}\\\\
ou\\\\ \sqrt[3]{5}](https://tex.z-dn.net/?f=5%5E%7B%5Cdfrac%7B3-1%7D%7B6%7D%7D%5C%5C%5C%5C+5%5E%7B%5Cdfrac%7B2%7D%7B6%7D%7D%5C%5C%5C%5C%0A5%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5C%0A+ou%5C%5C%5C%5C+%5Csqrt%5B3%5D%7B5%7D "5^{\dfrac{3-1}{6}}\\\\ 5^{\dfrac{2}{6}}\\\\
5^{\dfrac{1}{3}}\\\\
ou\\\\ \sqrt[3]{5}")
Letra C
(1+50)*50/2 = 1275
5+10+15+...+250 é uma PA de razão 5 e tem 50 termos. A soma é:
(5+250)*50/2 = 6375
Então fica:
(1275/6375) = 1/5
Letra C
Perguntas interessantes
Matemática,
11 meses atrás
Direito,
11 meses atrás
Ed. Física,
11 meses atrás
História,
1 ano atrás
História,
1 ano atrás
Matemática,
1 ano atrás