(1/2+1/6+1/12+...+1/72+1/90) : x = 9/20
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TD
26 tháng 1 2017
Ta có: \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{1}{10}-\frac{1}{9}-...-\frac{1}{2}-\frac{1}{1}\)
= \(\frac{9}{10}+\frac{1}{10}-\frac{1}{1}\)
= 1 - 1 = 0
Vậy kết quả của phép tính là 0
LN
1
29 tháng 6 2015
Ta có :
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
= 9/10 -( 1/90 + 1/72 + ... + 1/2)
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)}
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10)
= 9/10 - ( 1 - 1/10)
= 9/10 - 9/10
= 0
DQ
1
21 tháng 7 2015
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
=9/10-(1/9*10+1/8*9+...+1/1*2)
=9/10-(1/9-1/10+...+1-1/2)
=9/10-(-1/10+1)=9/10-9/10=0
BT
1
12 tháng 2 2020
= 9/1.10 + 1/9.10 + 1/8.9 + 1/7.8 + 1/6.7 +1/5.6 + 1/4.5 +1/3.4 +1/2.3 + 1/1.2
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 9/1.10 ( viết ngược lại)
= 1-1/2 + 1/2 -1/3 + 1/3 +....-1/10
= 1 - 1/10
= 9/10
HP
0
NT
4
25 tháng 6 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
25 tháng 6 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-...-\dfrac{1}{6}-\dfrac{1}{2}=-\left(-\dfrac{9}{10}+\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+...+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=-\left(-\dfrac{9}{10}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{10}\right)=-\left(-\dfrac{9}{10}+\dfrac{9}{10}\right)=0\)
15 tháng 3 2019
\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{9}{10.11}+\frac{1}{10}-1\)
\(=-\frac{9}{11}\)
16 tháng 8 2018
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
DD
Đoàn Đức Hà
Giáo viên
10 tháng 1 2021
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
giúp
=> (1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9 + 1/9.10) : x = 9/20
=> (1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9 + 1/9 - 1/10) : x = 9/20
=> (1 - 1/10) : x = 9/20
=> 9/10 : x = 9/20
X = 9/10 : 9/20 = 2