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NT
0
NT
1
17 tháng 7 2018
Ta có \(\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{12}\right)+...+\left(x-\frac{1}{90}\right)=1\)
\(\Rightarrow\left(x+x+x+...+x\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=1\)
\(\Rightarrow9x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)=1\)
\(\Rightarrow9x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)=1\)
\(\Rightarrow9x-\left(1-\frac{1}{10}\right)=1\)
\(\Rightarrow9x-\frac{9}{10}=1\)
\(\Rightarrow9x=\frac{19}{10}\)
\(\Rightarrow x=\frac{19}{10}\)
NT
0
12 tháng 7 2018
\(2\times x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(2\times x-2\times\frac{1}{12}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{10}\right)=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)+\frac{7}{30}=\frac{3}{11}\)
\(2\times\left(x-\frac{1}{12}\right)=\frac{13}{330}\)
\(x-\frac{1}{12}=\frac{13}{660}\)
\(x=\frac{17}{165}\)
12 tháng 7 2018
\(2x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{3}{11}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{3}{11}\)
\(\Rightarrow2x=\frac{3}{11}+\frac{2}{5}\)
\(\Rightarrow2x=\frac{37}{55}\)
\(\Rightarrow x=\frac{37}{55}:2\)
\(\Rightarrow x=\frac{37}{110}\)
Vậy \(x=\frac{37}{110}\)
_Chúc bạn học tốt_
QA
4
B1
30 tháng 7 2017
A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5)
+1/(5.6)+1/(6.7)+1/(7.8)
+1/(8.9)+1/(9.10)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6...
+1/9-1/10
=1-1/10
=9/10
thay x = a thôi đấy
chẳng động não gì cả
23 tháng 4 2016
a)12,8 x 195 + 12,8 x 804 + 12,8
= 12,8 x (195+804+1)
=12,8 x 1000
=12800
23 tháng 4 2016
a) = 12,8 \(\times\) ( 195 + 804 +1)
= 12,8 \(\times\) 1000
= 1280
b) = \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
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) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}=\dfrac{6}{x}\)
=>\(\dfrac{6}{x}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=>\(\dfrac{6}{x}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
=>\(x=6\times\dfrac{5}{2}=15\)