TÍNH TỔNG SAU A=32/8.11+32/11.14+...+32/1997.2000
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a) gọi d là UCLN ( n+3 ; 2n+5)
2n+5\(⋮\) d
n+3\(⋮\)d \(\Rightarrow\)2.(n+3) \(⋮\) d
vậy 2n+6 \(⋮\)d
2n+5 \(⋮\)d
=> 2n+6 - ( 2n+5) \(⋮\) d
=> 1\(⋮\) d
=> d =1
vì UCLN của n+3 và 2n+5 là 1 => dpcm
b)
\(A=\dfrac{3^2}{3}.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+....+\dfrac{1}{1997}-\dfrac{1}{2000}\right)\)
\(A=3.\left(\dfrac{1}{8}-\dfrac{1}{2000}\right)\)
\(A=\dfrac{747}{2000}\)
B dễ quá bạn suy nghĩ cho thông đầu nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)
x=0.78
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(x-\left(\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{1997.2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{1997.2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{1997}-\frac{1}{2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3\left(\frac{1}{8}-\frac{1}{2000}\right)=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-3.\frac{249}{2000}=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x-\frac{747}{2000}=\frac{-1}{2}\)
\(\Leftrightarrow\)\(x=\frac{-1}{2}+\frac{747}{2000}\)
\(\Leftrightarrow\)\(x=\frac{-253}{2000}\)
Vậy \(x=\frac{-253}{2000}\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(C=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)
\(C=3\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\right)\)
\(C=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(C=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(C=3.\frac{3}{25}\)
\(C=\frac{9}{25}\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
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=3(\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+...+\(\frac{3}{1997.2000}\))
=3(\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+...+\(\frac{1}{1997}\)-\(\frac{1}{2000}\))
=3(\(\frac{1}{8}\)-\(\frac{1}{2000}\))=3.\(\frac{249}{2000}\)=\(\frac{747}{2000}\)
\(A=3.\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{1997.2000}\right)\)
\(=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{1997}-\frac{1}{2000}\right)\)
\(=3.\left(\frac{1}{8}-\frac{1}{2000}\right)\)
\(=3.\frac{249}{2000}\)
\(\frac{747}{2000}\)