Question

Sa se efectueze:
$[( \frac{7}{45} + \frac{11}{30} -\frac{5}{12} ):1,0(5)-0,05]: \frac{1}{40} =?$

Answer 1

$[(\frac{7}{45}+ \frac{11}{30} - \frac{5}{12}):1,0(5)-0,05]: \frac{1}{40} =\\ \\1,0(5)= \frac{105-10}{90}= \frac{95}{90} = \frac{19}{18} ;0,05= \frac{5}{100}= \frac{1}{20}} \\ \\ 45= 3^{2}*5;30=2*3*5;12= 2^{2}*3\\ \\ (\frac{7}{45}+ \frac{11}{30} - \frac{5}{12})= \frac{2*7+6*11+15*5}{180} = \frac{183}{180} \\ \\ (\frac{183}{180}: \frac{19}{18} - \frac{1}{20} ): \frac{1}{40}$
$(\frac{183}{180}*\frac{18}{19} - \frac{1}{20} )*40= \frac{183*18-9*19}{180*19} *40= \\ \\ \frac{3294-171}{3420}*40= \\ \\ \frac{3123*2}{9}=694$
$[(\frac{7}{45}+ \frac{11}{30} - \frac{5}{12}):1,0(5)-0,05]: \frac{1}{40} =694$