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Đặt A=như đã cho.
=>1/2A=2/5*7+2/7*9+2/9*11+...+2/59*61.
=>1/2A=1/5-1/7+1/7-1/9+1/9-1/11+...+1/59-1/61.
=>1/2A=1/5-1/61=56/305.
=>A=56/305*2=112/305.
k nha đúng đó.Có j kb nha.
![](https://rs.olm.vn/images/avt/0.png?1311)
= 2/2 . ( 4 / 5.7 +4 / 7.9 +...+ 4 / 59.61 )
= 4/2 . ( 2 / 5.7 +2 / 7.9 +...+ 2 / 59.61 )
= 2 . ( 7-5 / 5.7 + 9-7 / 7.9 +...+ 61-59 / 59.61 )
= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 +...+ 1/59 - 1/61 )
= 2 . ( 1/5 - 1/61 )
= 2 . ( 61/305 - 5/305 )
= 2 . 56/305
= 112/305
ảnh anime đẹp thế là anime gì vậy bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{59.61}\)
\(\Rightarrow2.\left(\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{59.61}\right)\)
\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\right)\)
\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(\Rightarrow\frac{112}{305}\)
\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(=\frac{4.2}{5.7.2}+\frac{4.2}{7.9.2}+...+\frac{4.2}{59.61.2}\)
\(=\frac{4}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{4}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}_{ }\right)\)
\(=\frac{4}{2}.\left(\frac{1}{5}-\frac{1}{60}\right)\)
\(=\frac{4}{2}.\frac{11}{60}\)
\(=\frac{11}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(S=\frac{1}{5}-\frac{1}{61}\)
\(S=\frac{61}{305}-\frac{5}{305}\)
\(S=\frac{56}{305}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,Gọi tổng trên là A.
Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)
b,Gọi tổng trên là B
Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)
\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))
M= 3.(\(\frac{1}{5}-\frac{1}{61}\))
M=\(\frac{168}{305}\)
\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(M=\frac{84}{305}\)
phai la\(\frac{4}{7.9}\)chu ban
112/305