\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+\dfrac{3^2}{14.17}+....+\dfrac{3^2}{197.200}\)
\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)
=\(3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)
=\(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
=\(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
=\(3.\dfrac{3}{25}=\dfrac{9}{25}\)
Ta có: \(\dfrac{3^2}{8\cdot11}+\dfrac{3^2}{11\cdot14}+\dfrac{3^2}{14\cdot17}+...+\dfrac{3^2}{197\cdot200}\)
\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+...+\dfrac{3}{197\cdot200}\right)\)
\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)
Bài 6: Tính các tổng sau:
A= \(\dfrac{7}{10.11}\)+\(\dfrac{7}{11.12}\)+\(\dfrac{7}{12.13}\)+...+\(\dfrac{7}{69.70}\)
B= \(\dfrac{1}{25.27}\)+\(\dfrac{1}{27.29}\)+\(\dfrac{1}{29.31}\)+...+\(\dfrac{1}{73.75}\)
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
Tính giá trị
B=\(\dfrac{1}{25.27}\)+\(\dfrac{1}{27.29}\)\(\)+\(\dfrac{1}{29.31}\)+.....+\(\dfrac{1}{73.75}\)
`=>2B=(2)/(25.27)+(2)/(27.29)+(2)/(29.31)+....+(2)/(73.75)`
`=>2B=(1)/(25)-(1)/(27)+(1)/(27)-(1)/(29)+(1)/(29)-(1)/(31)+.....+(1)/(73)-(1)/(75)`
`=>2B=(1)/(25)-(1)/(75)`
`=>2B=(3)/(75)-(1)/(75)=(2)/(75)`
`=>B=(2)/(75):2`
`=>B=1/75`
\(B=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)
\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)
Ta có: \(B=\dfrac{1}{25\cdot27}+\dfrac{1}{27\cdot29}+\dfrac{1}{29\cdot31}+...+\dfrac{1}{73\cdot75}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{25\cdot27}+\dfrac{2}{27\cdot29}+\dfrac{2}{29\cdot31}+...+\dfrac{2}{73\cdot75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{75}=\dfrac{1}{75}\)
Thực hiện phép tính
a) A=\(\left(\dfrac{\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+.......+\dfrac{4}{399}}{\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+\dfrac{3^2}{14.17}+.....+\dfrac{3^2}{197.200}}\right).\dfrac{200720072007}{200820082008}\)
b) B=\(1.\sqrt{2}+2.\sqrt{3}+3.\sqrt{4}+....+9\sqrt{10}\)
c) D = \(\dfrac{2006}{0,20072008...}+\dfrac{2007}{0,020072008...}+\dfrac{2008}{0,0020072008}\)
\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)\(\dfrac{3}{11.14}\)+\(\dfrac{3}{14.17}\)<\(\dfrac{1}{2}\)
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
Tính:
\(\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)
Gọi \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)
là A, ta có
\(A=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)
\(\Rightarrow2.A=\dfrac{2}{25.27}+\dfrac{2}{27.29}+\dfrac{2}{29.31}+...+\dfrac{2}{73.75}\)\(\Rightarrow2.A=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2.A=\dfrac{1}{25}-\dfrac{1}{75}\)
\(\Rightarrow2.A=\dfrac{2}{75}\)
\(\Rightarrow A=\dfrac{2}{75}\div2\)
\(\Rightarrow A=\dfrac{1}{75}\)
KL: Vậy A =\(\dfrac{1}{75}\)
1. Tính
a. \(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+...+\dfrac{14}{197.200}\)
b. \(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
Bài 12: So sánh S = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{27.29}+\dfrac{2}{29.31}\)
P = \(\dfrac{2014}{2015}\)
\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{29\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ =\dfrac{1}{1}-\dfrac{1}{31}\\ =\dfrac{30}{31}\)
mà \(\dfrac{30}{31}>\dfrac{2014}{2015}\Rightarrow S>P\)
So sánh vs j nhỉ .-.?
`S=1-1/3+1/3-1/5+...+1/29-1/31`
`S=1-1/31=30/31`
S=2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/27-1/29+1/29-1/31)
S=2.(1-1/31)
S=2.30/31
S=60/31
P=2014/2015
=>S>P hay 60/31 > 2014 / 2015
