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\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
Bài 1 : Tính :
a, \(-\frac{8}{3}.\frac{6}{13}.\frac{7}{13}.\frac{-3}{8}+1\frac{3}{8}\)
\(=\left(-\frac{8}{3}.-\frac{3}{8}\right).\left(\frac{6}{13}.\frac{7}{13}\right)+1\frac{3}{8}\)
\(=1.\frac{42}{169}+1\frac{3}{8}\)
\(=\frac{2195}{1352}\)
b) \(75\%-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(-\frac{1}{2}\right)^2\)
\(=\frac{3}{4}-\frac{25}{6}+\frac{1}{4}\)
\(=\frac{3}{4}+\frac{1}{4}-\frac{25}{6}\)
\(=1-\frac{25}{6}\)
\(=-\frac{19}{6}\)
~Hok tốt~
Bài 2 :
a)\(\frac{3}{5}.x-\frac{2}{3}=\frac{1}{2}\) b) \(\left(\frac{1}{2}-x\right).\frac{2}{3}=\frac{1}{8}\)
\(\frac{3}{5}.x=\frac{1}{2}+\frac{2}{3}\) \(\frac{1}{2}-x=\frac{1}{8}:\frac{2}{3}\)
\(\frac{3}{5}.x=\frac{7}{6}\) \(\frac{1}{2}-x=\frac{3}{16}\)
\(x=\frac{7}{6}:\frac{3}{5}\) \(x=\frac{1}{2}-\frac{3}{16}\)
\(x=\frac{35}{18}\) \(x=\frac{5}{16}\)
c) \(\left|2x-\frac{3}{7}\right|-\frac{1}{2}=\frac{3}{4}\)
\(\left|2x-\frac{3}{7}\right|=\frac{3}{4}+\frac{1}{2}\)
\(\left|2x-\frac{3}{7}\right|=\frac{5}{4}\)
\(\Rightarrow2x-\frac{3}{7}\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(TH1:2x-\frac{3}{7}=\frac{5}{4}\) \(TH2:2x-\frac{3}{7}=-\frac{5}{4}\)
\(\Rightarrow2x=\frac{5}{4}+\frac{3}{7}\) \(\Rightarrow2x=-\frac{5}{4}+\frac{3}{7}\)
\(2x=\frac{47}{28}\) \(2x=-\frac{23}{28}\)
\(x=\frac{47}{28}:2\) \(2x=-\frac{23}{28}:2\)
\(x=\frac{47}{56}\) \(2x=-\frac{23}{56}\)
\(\Rightarrow x\in\left\{\frac{47}{56};-\frac{23}{56}\right\}\)
Câu 2:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}\)
\(=\frac{1.2.3...98}{2.3.4...99}\)
\(=\frac{1}{99}\)
k mình nha
ne ban con phan Tự Luan va phan Phan trac nghiem thi sao