Question

Sendo J=3+1/2.6/4-3/4 e R 165/32- (1+1/2).(2+1/2).(3/8+1), Calcule o valor da raiz quadrada de 3.J+R
Me ajudem!

Answer 1

$J=3 + \frac{1}{2}. \frac{6}{4}- \frac{3}{4}= 3+ \frac{3}{4}- \frac{3}{4} \\ J= 3 \\ R= \frac{165}{32}-(1+\frac{1}{2}).(2+ \frac{1}{2}).( \frac{3}{8}+1) \\ R=\frac{135}{32}-(\frac{3}{2}).(\frac{5}{2}).( \frac{11}{8})\\R= \frac{165}{32}- \frac{165}{8}= \frac{165 - 660}{32} \\R= \frac{-495}{32} \\ \sqrt{3.3 + (\frac{-495}{32}}) = \sqrt{\frac{288-495}{32}}= \sqrt{\frac{-207}{32}} \\ em,R,nao,existe,raiz$