dadas as dizimas periodicas 2,555... e 0,222... determine.
a) a soma delas, escrevendo resultados na forma abreviada.
b) O produto delas escrevendo o resultado ns forma de fração.
Soluções para a tarefa
Respondido por
303
Olá, Nilza.
![a)\ 2,555...+ 0,222.... = 2,777...=2+0,777...=2+\frac79=\\\\=\frac{18+7}9=\frac{25}9\\\\
b)\ 2,555...\times 0,222.... =(2+0,555...)\times0,222...=(2+\frac59)\times\frac29=\\\\=(\frac{18+5}9)\times\frac29=\frac{23}9\times\frac29=\frac{46}{81} a)\ 2,555...+ 0,222.... = 2,777...=2+0,777...=2+\frac79=\\\\=\frac{18+7}9=\frac{25}9\\\\
b)\ 2,555...\times 0,222.... =(2+0,555...)\times0,222...=(2+\frac59)\times\frac29=\\\\=(\frac{18+5}9)\times\frac29=\frac{23}9\times\frac29=\frac{46}{81}](https://tex.z-dn.net/?f=a%29%5C+2%2C555...%2B+0%2C222....+%3D+2%2C777...%3D2%2B0%2C777...%3D2%2B%5Cfrac79%3D%5C%5C%5C%5C%3D%5Cfrac%7B18%2B7%7D9%3D%5Cfrac%7B25%7D9%5C%5C%5C%5C%0Ab%29%5C+2%2C555...%5Ctimes+0%2C222....+%3D%282%2B0%2C555...%29%5Ctimes0%2C222...%3D%282%2B%5Cfrac59%29%5Ctimes%5Cfrac29%3D%5C%5C%5C%5C%3D%28%5Cfrac%7B18%2B5%7D9%29%5Ctimes%5Cfrac29%3D%5Cfrac%7B23%7D9%5Ctimes%5Cfrac29%3D%5Cfrac%7B46%7D%7B81%7D)
Respondido por
151
A) 2,555... + 0,222...=
= (2 + 5/9) + 2/9 =
18+5
= --------- + 2/9 =
9
= 23/9 + 2/9=
=25/9
b) 2,555... x 0,222...=
= 23/9 x 2/9 =
= 46/81
Espero ter ajudado. Abraços
= (2 + 5/9) + 2/9 =
18+5
= --------- + 2/9 =
9
= 23/9 + 2/9=
=25/9
b) 2,555... x 0,222...=
= 23/9 x 2/9 =
= 46/81
Espero ter ajudado. Abraços
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