cho A = \(\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
Chứng minh A = \(\frac{1}{101}\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\)
Những câu hỏi liên quan
1/ Cho Afrac{1}{1.102}+frac{1}{2.103}+...+frac{1}{299.400}Chứng minh rằng: Afrac{1}{101}left[left(1+frac{1}{2}+...+frac{1}{101}right)-left(frac{1}{300}+frac{1}{301}+...+frac{1}{400}frac{ }{ }right)right]2/ Tính Afrac{1}{2^2}+frac{1}{3^2}+...+frac{1}{2005^2}. Chứng minh A 13/ Cho Afrac{1}{101}+frac{1}{102}+...+frac{1}{200}Chứng minh: frac{1}{2} A 1GIÚP MÌNH NHA, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK AI NHANH NHẤT!!
Đọc tiếp
1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\frac{ }{ }\right)\right]\)
2/ Tính \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\). Chứng minh \(A< 1\)
3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Chứng minh: \(\frac{1}{2}< A< 1\)
GIÚP MÌNH NHA, MÌNH ĐANG CẦN GẤP.MÌNH SẼ TICK AI NHANH NHẤT!!
1.Cho A frac{1}{1.102}+frac{1}{2.103}+.....+frac{1}{299.400}Chứng minh rằng:Afrac{1}{101}[(1+frac{1}{2}+......+frac{1}{101})- (frac{1}{300}+frac{1}{301}+......+frac{1}{400})]
Đọc tiếp
1.Cho A =\(\frac{1}{1.102}\)+\(\frac{1}{2.103}\)+.....+\(\frac{1}{299.400}\)
Chứng minh rằng:
A=\(\frac{1}{101}\)\([\)\((\)1+\(\frac{1}{2}\)+......+\(\frac{1}{101}\)\()\)- \((\)\(\frac{1}{300}\)+\(\frac{1}{301}\)+......+\(\frac{1}{400})\)\(]\)
1/ Cho Afrac{1}{1.102}+frac{1}{2.203}+...+frac{1}{299.400}Chứng minh rằng: Afrac{1}{101}left[left(1+frac{1}{2}+...+frac{1}{101}right)-left(frac{1}{300}+frac{1}{301}+...+frac{1}{400}right)right]2/ Cho Afrac{1}{2^2}+frac{1}{3^2}+...+frac{1}{2005^2} Bfrac{1}{1.2}+frac{1}{2.3}+...+frac{1}{2004.2005}a) So sánh Avới B b) Chứng minh: A 13/ Cho Afrac{1}{101}+frac{1}{102}+...+frac{1}{200}Chứng minh: frac{1}{2} A 1 ...
Đọc tiếp
1/ Cho \(A=\frac{1}{1.102}+\frac{1}{2.203}+...+\frac{1}{299.400}\)
Chứng minh rằng: \(A=\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
2/ Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
a) So sánh \(A\)với \(B\) b) Chứng minh: \(A< 1\)
3/ Cho \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Chứng minh: \(\frac{1}{2}< A< 1\)
GIÚP MÌNH NHA, MÌNH CẢM ƠN. MÌNH ĐANG CẦN GẤP!!!!
2/
a, Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2005^2}< \frac{1}{2004.2005}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=B\)
b, \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}=1-\frac{1}{2005}< 1\)
3/
Ta có: \(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};...;\frac{1}{200}< \frac{1}{100}\)
\(\Rightarrow A< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\left(100ps\right)=\frac{1}{100}\cdot100=1\left(1\right)\)
Lại có: \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};...;\frac{1}{200}=\frac{1}{200}\)
\(\Rightarrow A>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\left(100ps\right)=\frac{1}{200}\cdot100=\frac{1}{2}\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{2}< A< 1\)
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vận dụng 3 A nha bn
dễ mà
xong tìm A ok nha bn
ok
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1/
\(A=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
=\(\frac{1}{101}\left(\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{101}{299.400}\right)\)
=\(\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{400}\right)\right]\)
=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)+\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+...+\frac{1}{299}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)
=\(\frac{1}{101}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{401}\right)\right]\)
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Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
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\(\frac{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{400}\right)}{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+\frac{1}{104}+...+\frac{1}{400}\right)}\)
G mk với, mk cần gấp lắm. Ai giải được mk k cho
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
Cho A= \(\dfrac{1}{1.102}+\dfrac{1}{2.103}+.....+\dfrac{1}{299.400}\)
Chứng minh rằng: A=\(\dfrac{1}{101}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\right]\)
Help me please.....
1/ Tính bằng cách thuận tiện nhất:Afrac{1}{2}+frac{5}{6}+frac{11}{12}+....+frac{9899}{9900}2/ Cho Afrac{1}{1.300}+frac{1}{2.301}+frac{1}{3.302}+....+frac{1}{101.400}Chứng minh rằng: Afrac{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!!!!!
Đọc tiếp
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)
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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/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
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Xem thêm câu trả lời
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
\(A=\frac{1}{1.300}+\frac{1}{2.301}+..........+\frac{1}{101.400}\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+........+\frac{299}{101.400}\)
\(\Rightarrow299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\Rightarrow299A=\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+.......+\frac{1}{400}\right)\)\(\Rightarrow\)\(A=\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)
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