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b) 10n+8=100..0+8=100...08 có tỏng các chữ số là 9 nên chia hết cho 9
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 :)))
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
\(=5^2\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}\right)\)
\(=25.\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{26}\right)\)
\(=5\left(1-\frac{1}{26}\right)\)
\(=5.\frac{25}{26}\)
\(=\frac{125}{26}\)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{1}{26.31}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(=1+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{11}+\frac{1}{11}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{31-1}{31}\)
\(=\frac{30}{31}\)
Vậy \(A=\frac{30}{31}\)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{31}\)
\(\Rightarrow A=\frac{30}{31}\)
Gọi A = 1/1.6 + 1/6.11 +...+ 1/(5n+1)(5n+6)
5A = 5/1.6 + 5/6.11 + ... + 5/(5n+1)(5n+6)
=1 - 1/6 + 1/6 - 1/11 + ... + 1/5n+1 - 1/5n+6
=1 - 1/5n+6 =5n+6/5n+6 - 1/5n+6=5n+5 /5n+6
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
\(B=\dfrac{10}{1.6}+\dfrac{10}{6.11}+\dfrac{10}{11.16}+.....+\dfrac{10}{46.51}\)
\(B=2\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+.....+\dfrac{1}{46}-\dfrac{1}{51}\right)\)
\(B=2\left(1-\dfrac{1}{51}\right)\)
\(B=2-\dfrac{2}{51}\)
\(B< 2\left(đpcm\right)\)