tính B = 3/1.4 + 5/4.9 + 7/9.16 + 15/16.31 + 19/31.50
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7 tháng 8 2016
\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{15}{16.31}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)
\(=1-\frac{1}{31}\)
\(=\frac{30}{31}\)
DS
7 tháng 8 2016
Dựa vào công thức được chứng minh:
(Em có thể chứng minh lại)
Ta có:
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}\)
\(=1-\frac{1}{31}\)
\(=\frac{30}{31}\)
Chúc em học tốt^^
18 tháng 9 2021
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
1 tháng 11 2018
\(\text{Ta có :}\)
\(\frac{3}{1.4}=1-\frac{1}{4}\)
\(\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\)
\(\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\)
\(......\)
\(\frac{19}{81.100}=\frac{1}{81}-\frac{1}{100}\)
\(\text{Cộng vế với vế ta có:}\)
\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Ko biết có đc k ko ta!?
VT
0
IM
1 tháng 8 2016
a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
YN
21 tháng 2 2017
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)
TT
29 tháng 7 2016
Thanks bạn nha , đúng là trường hợp khẩn cấp thì lại có bạn , thanks bạn so ,so ,so, so ,so,.... much
14 tháng 4 2022
a: \(\dfrac{3}{5}-\dfrac{7}{4}=\dfrac{12}{20}-\dfrac{35}{20}=\dfrac{-23}{20}\)
b: \(\dfrac{4}{5}:\dfrac{-8}{15}=\dfrac{-4}{5}\cdot\dfrac{15}{8}=\dfrac{-60}{40}=\dfrac{-3}{2}\)
c: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
d: \(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)+\dfrac{5}{7}=\dfrac{2}{7}\)
B = 3/1.4 + 5/4.9 + 7/9.16 + 15/16.31 + 19/31.50
Ta co cong thuc : 1/n.(n-1)=1/n-1/n-1
=> B = 1/1 - 1/4 +1/4 - 1/9 + 1/9 - 1/16 + 1/16 - 1/31 + 1/31 - 1/50
= 1/1 - 1/50
= 49/50