\(\left(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\right)\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
Ta có :
\(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{53.55}\)
\(=\frac{4}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\left(\frac{11}{55}-\frac{1}{55}\right)=2.\frac{10}{55}=2.\frac{2}{11}=\frac{4}{11}\)
Có gì không hiểu bạn hỏi lại mình nhé! Chúc bạn học tốt!
Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{53.55}\)
Đặt C = \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{53.55}\)
\(\frac{1}{2}C=\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+....+\left(\frac{1}{53}-\frac{1}{55}\right)\)
\(\frac{1}{2}C=\frac{1}{5}-\frac{1}{55}\)
\(\frac{1}{2}C=\frac{2}{11}\)
\(C=\frac{2}{11}:\frac{1}{2}\)
Vậy C = \(\frac{4}{11}\)
Có gì sai thì mong bạn thông cảm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\frac{402}{2015}\)
\(=\frac{603}{2015}\)
b)\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\cdot\frac{95}{294}\)
\(=\frac{38}{147}\)
a) Gọi tổng trên là A
A = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)
A == \(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+\frac{3}{9}-\frac{3}{11}+...+\frac{3}{2013}-\frac{3}{2015}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
A = \(\frac{3}{5}-\frac{3}{2015}\)
A = \(\frac{1209}{2015}-\frac{3}{2015}\)
A = \(\frac{1206}{2015}\)
b) Gọi tổng trên là B
B = \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)
B = \(\frac{4}{3}-\frac{4}{8}+\frac{4}{8}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{93}-\frac{4}{98}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
B = \(\frac{4}{3}-\frac{4}{98}\)
B = \(\frac{686}{294}-\frac{12}{294}\)
B = \(\frac{674}{294}=\frac{337}{147}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
![](https://rs.olm.vn/images/avt/0.png?1311)
2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5
2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5
2\3x-780\11:[13\2.(1\3-1\11)]=-5
2\3x-780\11:[13\2.8\33]=-5
2\3x-780\11:52\33=-5
2\3x-525\13=-5
2\3x=-5+525\13
2\3x=460\13
x=460\13:2\3
x=690\13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)
<=> \(2.\frac{1}{2}-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=\frac{8}{5}-2x\)
<=> \(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)=\frac{8}{5}-2x\)
<=> \(1-\left(1-\frac{1}{11}\right)-\frac{8}{5}=-2x\)
<=> \(-\frac{83}{55}=-2x\)
<=> \(x=\frac{83}{110}\)
Ta có :
\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+............+\frac{2}{53.55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..............+\frac{1}{53}-\frac{1}{55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\frac{2}{11}=\frac{4}{11}\)
k nha bạn !!!