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8) (1+frac(1)/(2))(3-(5)/(2)) Cdot Frac(3,5){}=

Question

8) (1+frac(1)/(2))(3-(5)/(2)) cdot frac(3,5){}=

Solution

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4.4 (311 Votos)
Zoraide Avançado · Tutor por 1 anos

Resposta

Para resolver essa expressão, vamos seguir a ordem das operações matemáticas.Primeiro, vamos simplificar a fração :\( \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} ={1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{1}{2}}{3-\frac{5}{2}} = \frac{1+\frac{