Cho: A= \(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
Chứng minh rằng:
\(A=\frac{1}{299}\)(\(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\))
Cho A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\)
CMR:A=\(\frac{1}{299}.\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.....+\frac{1}{400}\right)\)
ở chỗ 1/299 là nhân với ngoặc vuông nha bạn nào giải hộ mình rthì i li-ke
\(A=\frac{1}{1.300}+\frac{1}{2.301}+..........+\frac{1}{101.400}\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+........+\frac{299}{101.400}\)
\(\Rightarrow299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\Rightarrow299A=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.......+\frac{1}{400}\right)\)\(\Rightarrow\)\(A=\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)
1/ Tính bằng cách thuận tiện nhất:
\(A=\)\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{9899}{9900}\)
2/ Cho \(A=\)\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\)
Chứng minh rằng: \(A=\)\(\frac{1}{299}\).\(\left[\left(1+\frac{1}{2}+....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\right]\)
GIÚP MÌNH VS, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK CHO AI NHANH NHẤT!!!!!
Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)
Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)
1/
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)
2/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
Cho A = \(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+.....+\frac{1}{101.400}\)
Chứng minh rằng : \(A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)\right]-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\)
Mình đang cần gấp lắm từ giờ tới 1h30' chiều nhé ! Ai làm xong mà có cách làm cho hẳn 3 tick ( ko cần biết đúng hay sai)
\(A=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.......+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+....+\frac{1}{400}\right)\right]\)
=>đpcm
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{.302}+....+\frac{1}{101.400}\)
=> \(299.A=\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}\)
=> \(299.A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
=> \(A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
Có j không hiểu có thể hỏi lại mk
Chúc bạn làm bài tốt
\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+\frac{1}{3\times302}+...+\frac{1}{100\times399}+\frac{1}{101\times400}\)
\(A=\frac{1}{299}\times\left(\frac{299}{1\times300}+\frac{299}{2\times301}+\frac{299}{3\times302}+...+\frac{299}{100\times399}+\frac{299}{101\times400}\right)\)
\(A=\frac{1}{299}\times\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{100}-\frac{1}{399}+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{399}+\frac{1}{400}\right)\right]\)
=> đpcm
Cho A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\);
B=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{299.400}\)
Lưu ý dấu chấm là dấu nhân nhé
Tính \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
có học thì mới có ăn không làm mà đòi có điểm chỉ có ăn ...
Tính: \(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}}\)
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ngu vậy chết mm đi
\(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+.............+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+.......+\frac{1}{299.400}}\)
Tính:
A=\(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}}\)
Giúp mình nha các bạn
cho \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+.....+\frac{1}{101.400}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+....+\frac{1}{299.400}\)
so sánh A và B