Question

bài 1:

a) 10\17+5\-13-11\25+7\17-8\13

b) 0,3-93\7-70%0-4\7

c)1\8+1\24+1\48+...+1\2400

d) 3\41-12\17+33\49

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12\41-48\17+132\49

Answer 1

a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)

= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)

= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)

= 1 - 1 - \(\dfrac{11}{25}\)

= - \(\dfrac{11}{25}\)

Answer 2

b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)

= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))

= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)

= - \(\dfrac{499}{35}\)

Answer 3

c, \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{48}\)+ ...+ \(\dfrac{1}{2400}\)

= \(\dfrac{1}{2\times4}\)  +\(\dfrac{1}{4\times6}\) + \(\dfrac{1}{6\times8}\) +...+ \(\dfrac{1}{48\times50}\)

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{2\times4}\) +\(\dfrac{2}{4\times6}\)+ \(\dfrac{2}{6\times8}\) +...+ \(\dfrac{2}{48\times50}\))

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{48}\) - \(\dfrac{1}{50}\))

= \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{12}{25}\)

= \(\dfrac{6}{25}\)

Answer 4

d, D = \(\dfrac{\dfrac{3}{41}-\dfrac{12}{17}+\dfrac{33}{49}}{\dfrac{12}{41}-\dfrac{48}{17}+\dfrac{132}{49}}\)

   D = \(\dfrac{\dfrac{3}{41}-\dfrac{12}{17}+\dfrac{33}{49}}{4\times\left(\dfrac{3}{41}-\dfrac{12}{17}+\dfrac{33}{49}\right)}\)

D = \(\dfrac{1}{4}\)