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t=(1-1/2+1-1/6+1-1/12+..........+1-1/90+1-1/110)-10/11
t=(1+1+1+...........+1+1)-(1/2+1/6+1/12+.......+1/90+1/110)-10/11
có 10 số 1 vì từ 1/2 đến 109/110 có 10 p/số
t=1.10-(1/1.2+1/2.3+1/3.4+.........+1/9.10+1/10.11)-10/11
t=10-(1-1/2+1/2_1/3+1/3_1/4+.............+1/9-1/10+1/10-1/11)-10/11
t=10-(1-1/11)-10/11
t=10-10/11-10/11
t=10-(10/11+10/11)
t=10-10/11.2
t=10-20/11
t=110/11-20/11
t=90/11
T = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}+\frac{10}{11}\)
T = \(\frac{10}{11}+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{110}\right)\)
T = \(\frac{10}{11}+\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
T = \(\frac{10}{11}+10+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
T = \(\frac{120}{11}+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
T = \(\frac{120}{11}-\left(1-\frac{1}{11}\right)\)
T = \(\frac{120}{11}-\frac{9}{11}\)
T = \(\frac{111}{11}\)
bài của bạn Giang tính nhầm bước thứ hai từ cuối lên
Sửa: T = \(\frac{120}{11}-\frac{10}{11}=10\)
\(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)
\(=\frac{12-1}{12}+\frac{20-1}{20}+\frac{30-1}{30}+\frac{42-1}{42}+\frac{56-1}{56}+\frac{72-1}{72}+\frac{90-1}{90}+\frac{110-1}{110}\)
\(=1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=8-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{256}{33}\)
ĐẶT A = \(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{109}{110}\)
8 -A = \(1-\frac{11}{12}+1-\frac{19}{20}+1-\frac{29}{30}+1-\frac{41}{42}+1-\frac{55}{56}+1-\frac{71}{72}+1-\frac{89}{90}+1-\frac{109}{110}\)
8 -A \(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
8 -A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
8 - A = \(\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
=> A = \(8-\frac{8}{33}=\frac{256}{33}\)
\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=8+\frac{1}{10}=\frac{81}{10}\)
\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
=\(9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=8+\frac{1}{10}=\frac{81}{10}\)
\(=8+\frac{1}{10}=\frac{81}{10}\)
Ta có : \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=\frac{81}{10}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)
\(=(1-\frac{1}{2})+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\) \(+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\frac{11}{10}\)
\(=\frac{79}{10}\)
~Học tốt nha~
Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(\Leftrightarrow A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+......+\left(1-\frac{1}{90}\right)\)
\(\Leftrightarrow A=\left(1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{90}\right)\)
\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(\Leftrightarrow A=9-\left(1-\frac{1}{10}\right)\)
\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)
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