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\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}\)
\(A=\frac{1}{100}\)
\(B=1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{72}\)
\(B=1+1+...+1+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}\right)\)
\(B=5.1+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(B=5+\frac{2}{9}=\frac{47}{9}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(=\frac{1.2.3.4....99}{2.3.4.5...100}\)
\(=\frac{1}{100}\)
Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
\(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)
\(=(1-\frac{1}{2})+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\) \(+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\frac{11}{10}\)
\(=\frac{79}{10}\)
~Học tốt nha~
Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(\Leftrightarrow A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+......+\left(1-\frac{1}{90}\right)\)
\(\Leftrightarrow A=\left(1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{90}\right)\)
\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(\Leftrightarrow A=9-\left(1-\frac{1}{10}\right)\)
\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)=9-1+\frac{1}{10}=8\frac{1}{10}\)
=> A = ( 1 - 1/2 ) + ( 1 - 1/6 ) + ( 1 - 1/12 ) + ( 1 - 1/30 ) + .... + ( 1 - 1/90 )
=> A = ( 1 + 1 + 1 + .... 1 ) - ( 1/2 + 1/6 + 1/12 + 1/30 + .... + 1/90 )
=> A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/9.10 )
=> A = 9 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10 )
=> A = 9 - ( 1 - 1/10 )
=> A = 9 - 9/10
=> 81/10
A=1-1/2+1-1/6+...+1-1/90
=9-(1/2+1/6+...+1/90) =9-(1/1.2+1/2.3+...+1/9.10)
=9-(1-1/10)=9-9/10=81/10
bn vào câu hỏi tương tự sẽ có chi tiết . Nếu k thì bn hãy để ý mỗi tử đều bé hơn mẫu 1 đơn vị sau đó bn tách ra bằng cách lấy 1 trừ . VD: 5/6 bằng 1 - 1/6 . Đến đó đếm đc 9 chữ số 1 ta lấy 9 làm sbt trừ đi tổng của các ps ta tách đc . Khi đó thì bài toán quá đơn giản rồi . Chúc bn học tốt
(1-1/2)+(1-1/6)+...+(1-1/90)
9+(1/2+1/6+...+1/90)
9+(1/1.2+1/2.3+...+1/9.10)
9+1-9/10=9/1/10=91/10
