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Số số hạng của dãy: (2012-2):1+1=2011(số)
Vì 2011 là số lẻ nên A là số âm.
a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)
b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)
c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)
d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
Bài 1:
a, \(\dfrac{-x-2}{3}\) = - \(\dfrac{6}{7}\)
- \(x\) - 2 = - \(\dfrac{18}{7}\)
\(x\) = - 2 + \(\dfrac{18}{7}\)
\(x\) = - \(\dfrac{4}{7}\)
Bài b, \(\dfrac{4}{7-x}\) = \(\dfrac{1}{3}\)
12 = 7 - \(x\)
\(x\) = 7 - 12
\(x\) = -5
\(B=\frac{\left(\frac{1}{5}-\frac{2}{7}\right).\frac{3}{4}-\frac{3}{4}.\left(\frac{1}{3}-\frac{2}{7}\right)}{\frac{1}{5}.\frac{2}{7}-\frac{1}{3}.\left(\frac{2}{7}+\frac{3}{9}\right)+\frac{3}{9}.\frac{1}{5}}\)
\(B=\frac{\frac{3}{4}.\left(\frac{1}{5}-\frac{2}{7}-\frac{1}{3}+\frac{2}{7}\right)}{\frac{1}{5}.\frac{2}{7}-\frac{1}{3}.\frac{2}{7}-\frac{1}{3}.\frac{1}{3}+\frac{1}{3}.\frac{1}{5}}\)
\(B=\frac{\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{3}\right)}{\frac{2}{7}.\left(\frac{1}{5}-\frac{1}{3}\right)-\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{5}\right)}\)
\(B=\frac{\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{3}\right)}{\frac{2}{7}.\left(\frac{1}{5}-\frac{1}{3}\right)+\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{3}\right)}\)
\(B=\frac{\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{3}\right)}{\left(\frac{2}{7}+\frac{1}{3}\right).\left(\frac{1}{5}-\frac{1}{3}\right)}\)
\(B=\frac{\frac{3}{4}.\left(\frac{1}{5}-\frac{1}{3}\right)}{\frac{13}{21}.\left(\frac{1}{5}-\frac{1}{3}\right)}=\frac{\frac{3}{4}}{\frac{13}{21}}=\frac{3}{4}:\frac{13}{21}=\frac{63}{52}\)