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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{26}-\frac{1}{31}\)
\(\Rightarrow A=1-\frac{1}{31}\)
\(\Rightarrow A=\frac{30}{31}\)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(=1+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{11}+\frac{1}{11}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{31-1}{31}=\frac{30}{31}\)
\(\text{Vậy }A=\frac{30}{31}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
B = 10/1.6 + 10/6.11 + 10/11.16 + ... + 10/46.51
B = 2(5/1.6 + 5/6.11 + 5/11.16 + ... + 5/46.51)
B = 2 . (1 - 1/6 + 1/6 - 1/11 + 1/11 - 1/16 + ... + 1/46 - 1/51)
B = 2. (1 - 1/51)
B = 2.50/51
B = 100/51 < 102/51 = 2
=> Ta có đpcm
\(B=\dfrac{10}{1.6}+\dfrac{10}{6.11}+\dfrac{10}{11.16}+.....+\dfrac{10}{46.51}\)
\(B=2\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+.....+\dfrac{1}{46}-\dfrac{1}{51}\right)\)
\(B=2\left(1-\dfrac{1}{51}\right)\)
\(B=2-\dfrac{2}{51}\)
\(B< 2\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{1}{26.31}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(=1+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{11}+\frac{1}{11}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{31-1}{31}\)
\(=\frac{30}{31}\)
Vậy \(A=\frac{30}{31}\)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{31}\)
\(\Rightarrow A=\frac{30}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(=5\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{31-26}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
đặt biểu thức là A
5A=(1/1x6+1/6x11+...+1/96x101)x5=5/1x6+5/6x11+...+5/96x101
5A=6-1/1x6+11-6/6x11+...+101-96/96x101
5A=6/1x6-1/1x6+11/6x11-6/6x11+...+101/96x101-96/96x101
5A=1-1/6+1/6-1/11+...+1/96-1/101(sau khi rút gọn các phân số)
5A=1-1/101(còn lại sau khi trừ)
5A=100/101
A=100/101:5=20/101
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{25}{1.6}+\frac{25}{6.11}+........+\frac{25}{86.91}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+.....+\frac{5}{86.91}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+......+\frac{1}{86}-\frac{1}{91}\right)\)
\(=5.\left(1-\frac{1}{91}\right)\)
\(=\frac{90}{91}\)
gọi đề cho = A
có : A= 5.( (5/1.6) + (5/6.11) + (5/11.16) + ... + (5/101.106)
A = 5. (5 - 5/106) ( theo tính chất)
từ đây bạn có thể tính A nhé :)))