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\(\dfrac{5}{12}+\dfrac{1}{3.8}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}+\dfrac{1}{24}-\dfrac{1}{6}\)
\(=\dfrac{10}{24}+\dfrac{1}{24}-\dfrac{4}{24}\)
\(=\dfrac{11}{24}-\dfrac{4}{24}\)
\(=\dfrac{7}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\)
\(=1-\dfrac{1}{40}\)
\(=\dfrac{39}{40}\)
1/(1×4) + 1/(4×7) + ... + 1/(34×37) + 1/(37×40)
= 1/3 × (1 - 1/4 + 1/4 - 1/7 + ... + 1/34 - 1/37 + 1/37 - 1/40)
= 1/3 × (1 - 1/40)
= 1/3 × 39/40
= 13/40
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\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)
\(=1-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=1-\dfrac{1}{40}\)
\(=\dfrac{39}{40}\)
Vậy giá trị cần tìm là: \(\dfrac{39}{40}\)
\(3xA=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+...+\dfrac{37-34}{34x37}+\dfrac{40-37}{37x40}=\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{17}{ }+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}=\)
\(=1-\dfrac{1}{40}=\dfrac{39}{40}\Rightarrow A=\dfrac{39}{40}:3=\dfrac{13}{40}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{100\times103}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{100\times103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{34}{103}\)