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bài A: áp dụng công thức: 1 + 2 + 3 + ... + n = n x (n + 1) : 2 tính được 5050
bài B: áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rồi triệt tiêu gần hết, qui đồng mẫu số tính được B = 99/100
A = 1 + 2 + 3 + 4 + 5 + ... + 99 + 100
= ( 100 + 1 ) x 100 : 2 = 5050
Vậy A = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy \(B=\frac{99}{100}\)
Học tốt #
C=\(\frac{7}{3.4}\)-\(\frac{9}{4.5}\)+\(\frac{11}{5.6}\)+\(\frac{13}{6.7}\)+\(\frac{15}{7.8}\)-\(\frac{17}{8.9}\)+\(\frac{19}{9.10}\)
=\(\frac{1}{3}\)+\(\frac{1}{4}\)-\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{6}\)-\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)+\(\frac{1}{8}\)-\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)+\(\frac{1}{10}\)
=\(\frac{1}{3}\)+\(\frac{1}{10}\)=\(\frac{13}{30}\)
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{19.20}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{5}-\frac{1}{20}\)
\(=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
Đặt \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)
\(A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{18-16}{16.18}\)
\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{18}\right)\)
\(A=\frac{4}{2}.\frac{4}{9}\)
\(\Rightarrow A=\frac{8}{9}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)
\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{18}\right)\)
\(=2.\frac{4}{9}\)
\(=\frac{8}{9}\)
\(A=\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}+...+\frac{4}{47.48}\)
\(A=4.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+......+\frac{1}{47.48}\right)\)
\(A=4.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{47}-\frac{1}{48}\right)\)
\(A=4.\left(\frac{1}{4}-\frac{1}{48}\right)\)
\(A=4.\frac{11}{48}\)
\(A=\frac{11}{12}\)