\(A=\frac{2}{4.7} +\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{73.76}\)
Những câu hỏi liên quan
\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+.....+\frac{1}{73.76}\)
A=\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)
3A=3(\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\))
3A=\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
3A=\(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}+\frac{1}{76}\)
3A=\(\frac{1}{4}-\frac{1}{76}\)
3A=\(\frac{9}{38}\)
A=\(\frac{9}{38}\):3
A=\(\frac{3}{38}\)
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đặt A=1/4.7+1/7.10+...+1/73.76
3A=1/4-1/7+1/7-1/10+...+1/ 73 -1/ 76
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\(A=\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\)
\(3A=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(3A=\frac{1}{4}-\frac{1}{76}\)
\(3A=\frac{9}{38}\)
\(A=\frac{3}{38}\)
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Xem thêm câu trả lời
tính nhanh : A = \(\frac{2}{2.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{37.39}\)
B = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+......+\frac{1}{73.76}\)
http://olm.vn/hoi-dap/question/772291.html
sau 3 phút có kết quả tùy bạn
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A=\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{73.76}\)
Hãy tính A
C=\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+\(\frac{3}{10.13}\)+.......+\(\frac{3}{73.76}\)
C = 3/4.7 + 3/7.10 + 3/10.13 + ... + 3/73.76
=1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/73 - 1/76
=1/4 - 1/76
=18/76
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\(C=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+......+\frac{1}{73}-\frac{1}{76}\)
\(=\frac{1}{4}-\frac{1}{76}\)
\(=\frac{19}{76}-\frac{1}{76}\)
\(=\frac{18}{76}=\frac{9}{38}\)
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\(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(C=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(C=\frac{1}{4}-\frac{1}{76}\)
\(C=\frac{9}{38}\)
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Xem thêm câu trả lời
Tìm A :
A = \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)
A:3=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\)
A:3=\(\frac{1}{1}-\frac{1}{100}\)
A:3=\(\frac{99}{100}\)
A=\(\frac{99}{100}.3\)
A=\(\frac{297}{100}\)
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\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(A:3=\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A:3=\frac{1}{1}-\frac{1}{100}\)
\(A:3=\frac{99}{100}\)
\(A=\frac{99}{100}.3\)
\(A=\frac{297}{100}\)
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Tính :
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Ta thấy :
\(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
\(.........\)
\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)
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đáp án = \(\frac{297}{100}\)
đúng không?
kết bạn với mh nha
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Bài 1a)frac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{99.100} b)frac{2}{3.5}+frac{2}{5.7}+frac{2}{7.9}+...+frac{2}{2015.2017}c)left(frac{3}{2}+frac{7}{6}+frac{13}{12}+...+frac{9901}{9900}right)-frac{99}{100} d)frac{3}{4.7}+frac{3}{7.10}+frac{3}{10.13}+...+frac{3}{73.76}
Đọc tiếp
Bài 1
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\)\(\frac{1}{99.100}\) b)\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\)
c)\(\left(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{9901}{9900}\right)\)\(-\frac{99}{100}\) d)\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
a, 1/1.2+1/1.3+...+1/99.100
= 1-1/2+1/2-1/3+1/3+...+1/99-1/100
=1-1/100
=99/100
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A = 99/100
B = 2014/6051
C = 99
D = 9/38
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Tính tổng:
A=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
B=\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(A=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{97}{303}=\frac{97}{606}\)
\(B=\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+...+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{100\cdot103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(\frac{1}{4}-\frac{1}{103}\right)=\frac{2}{3}\cdot\frac{99}{412}=\frac{33}{206}\)
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\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+...+\dfrac{2}{73.76}\)
help me,sáng mai thầy kiểm tra bài của mình
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Ta có : A = 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 .
⇒ 3/2 A = 3/2 . ( 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 ) .
⇒ 3/2 A = 3/ 4.7 + 3/ 7.10 + ... + 3/ 73.76 .
= 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/ 73 - 1/76 .
= 1/4 - 1/76 .
= 19/76 - 1/76 .
= 9/38 .
Do đó : A = 9/38 : 3/2 .
= 9/38 . 2/3 .
= 3/19 .
Vậy A = 3/19 .
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