Tìm số tự nhiên x thỏa mãn:
a) 5/1.6 + 5/6.11 +...+ 5/(5x + 1).(5x + 6) = 2010/2011
b) 1/3.5 + 1/5.7 + 1/7.9 +...+ 1/(2x + 1).(2x + 3) = 15/93
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TA
10 tháng 8 2015
1-1/6+1/6-1/11+....+1/(5x+1)-1/(5x+2)=2010/2011 <=>1-1/(5x+2)=2010/2011 <=>1/2011=1/(5x+2) <=>x=401
SK
7
XO
4 tháng 8 2020
Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
H
1
15 tháng 4 2015
Ta có:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}\)
\(=1-\frac{1}{5x+6}\)
\(=\frac{5x+5}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow5x+5=2010\)
\(\Rightarrow5x=2010-5=2005\)
\(\Rightarrow x=2005:5=401\)
Vậy x=401
HP
1
10 tháng 8 2018
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
17 tháng 3 2017
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
