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\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{59\cdot60}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{69}-\frac{1}{60}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
1) \(\frac{1}{5}-\frac{1}{6}=\frac{6-5}{5.6}=\frac{1}{5.6}\)
\(\frac{1}{6}-\frac{1}{7}=\frac{7-6}{6.7}=\frac{1}{6.7}\)
2) Áp dụng bài trên, ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
= \(1-\frac{1}{6}=\frac{5}{6}\)
Gọi \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{17.18}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{17}-\frac{1}{18}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{17}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{18}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{18}-1-\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{9}\)
\(=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{18}\)
Ta thấy : \(\frac{1}{10}>\frac{1}{19};\frac{1}{11}>\frac{1}{19};\frac{1}{12}>\frac{1}{19};....;\frac{1}{18}>\frac{1}{19}\)
\(\Rightarrow A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{18}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\)(có 9 số \(\frac{1}{19}\) )
\(\Rightarrow A>9.\frac{1}{19}=\frac{9}{19}\)(đpcm)
Bạn muốn có đáp án mk chỉ cho bạn hai cách:
1, Vô âu hỏi tương tự.
2, lên google gõ dòng chữ Cho p=1/1.2+1/3.4+1/5.6+...+1/99.100. Chứng minh 7/12<P<5/6 rồi nhấn tìm kiếm quá trời câu trả lời luôn á .
Mk lười giải quá bạn làm theo cách mk chỉ đi thể nào cũng có câu trả lời mà tìm được câu trả lời thì nhớ k cho mk nhé ....
1/1 . 2 + 1/ 3 . 4 + 1/5 . 6 + ...+ 1/99 . 100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/99 - 1/100
= ( 1 + 1/3 + 1/5 + ...+ 1/99 ) - ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - 2 . ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - ( 1 + 1/2 + ...+ 1/50 )
= 1/51 + 1/52 + ...+ 1/100
Tham khảo nha !!!
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) (đpcm)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
Ta có : \(\frac{5.5}{1.6}+\frac{5.5}{6.11}+\frac{5.5}{11.16}+\frac{5.5}{16.21}\)
\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)\)
\(=5\left(1-\frac{1}{21}\right)\)
\(=5.\frac{20}{21}=\frac{100}{21}\)
= 1 -1/2 + 1/2 - 1/3 + 1/3 -...-1/60
= 1-1/60=59/60