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a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))
=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
=\(\frac{250}{101}\)
\(=\frac{5}{2}.\frac{100}{101}\)
a,21.321.3+23.523.5+25.725.7+....+299.101
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=>\(\frac{1}{1}-\frac{1}{101}\)
=>\(\frac{100}{101}\)
b,
51.351.3+53.553.5+55.755.7+....+599.101
=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)
=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}.\frac{100}{101}\)
=>\(\frac{250}{101}\)
\(B=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)=\)\(\frac{250}{51}\)
\(B=5\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5\left(\frac{51}{51}-\frac{1}{51}\right)=\frac{250}{51}\)
\(B=5.\frac{50}{51}=\frac{250}{51}\)
\(B=\frac{250}{51}=\frac{250}{51}\)
\(a,x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=--\frac{37}{45}.\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}=\frac{37}{45}+\frac{1}{45}=\frac{38}{45}\)
\(x=\frac{38}{45}-\frac{1}{5}=\frac{29}{45}\)
\(b,\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Rightarrow5x+6=2016\)
\(\Rightarrow5x=2010\Rightarrow x=402\)
\(c,\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow x+2=2018\Rightarrow x=2016\)
học tốt ~~~
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{6}{31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
Ta có :
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
\(S=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
\(S=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(S=5\left(1-\frac{1}{26}\right)\)
\(S=5.\frac{25}{26}\)
\(S=\frac{125}{26}\)
Vậy \(S=\frac{125}{26}\)
Chúc bạn học tốt ~
\(S=\frac{5^2}{1\cdot6}\cdot\frac{5^2}{6\cdot11}\cdot\frac{5^2}{11\cdot16}\cdot\frac{5^2}{16\cdot21}\cdot\frac{5^2}{21\cdot26}\)
\(=5\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{5}{16\cdot21}+\frac{5}{21\cdot26}\right)\)
\(=5\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(=5\cdot\left(1-\frac{1}{26}\right)\)
\(=5\cdot\frac{25}{26}\)
\(=\frac{125}{26}\)
ta co \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
=\(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+\frac{5}{26.31}\right)\)
=\(5.\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}+\frac{26-21}{21.26}+\frac{31-26}{26.31}\right)\)
=\(5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
=\(5.\left(1-\frac{1}{31}\right)\)
=\(5.\frac{30}{31}\)
=\(\frac{150}{31}\)
a) 2A= 1+1/2^2+1/2^3+...+1/2^2015+1/2^2016
2A-A=(1+1/2+1/2^2+...+1/2^2015+1/2^2016)-(1/2+1/2^2+...+1/2^2016+1/2^2017)
A= 1-1/2^2017
b) B=5.(5/1.6+5/6.11+...+5/26.31)
B=5.(1/5-1/6+1/6-1/11+1/11...-1/26+1/26-1/31)
B= 5.(1/5-1/31)
B=5.26/155
B=26/31
Đề bài của bạn sai rồi đề đúng phải là 2/1.3 + 5/3.5 + 5/5.7 + ... + 5/101.103
a=5/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/101.103) = 5/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/101 - 1/103)
= 5/2 .(1 - 1/103) = 5/2 . 102/103 = 255/103
câu b tương tự. nhưng đề bài b của bạn cũng sai rồi : 2/1.6 + 2/6.11 + 2/11.16 + ... + 2/101.106