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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
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
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
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
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
Bài 3:
a: \(\frac13+\frac23:x=-7\)
=>\(\frac23:x=-7-\frac13=-\frac{22}{3}\)
=>\(x=\frac23:\frac{-22}{3}=\frac{-2}{22}=-\frac{1}{11}\)
b: \(3\frac12-\frac12x=\frac23\)
=>\(\frac72-\frac{x}{2}=\frac13\)
=>\(7-x=\frac23\)
=>\(x=7-\frac23=\frac{21}{3}-\frac23=\frac{19}{3}\)
c: \(\left\lbrack\left(x+\frac13\right)\cdot\frac34+5\right\rbrack:2=3\)
=>\(\left(x+\frac13\right)\cdot\frac34+5=3\cdot2=6\)
=>\(\left(x+\frac13\right)\cdot\frac34=1\)
=>\(x+\frac13=1:\frac34=\frac43\)
=>\(x=\frac43-\frac13=\frac33=1\)
d: \(\left(1\frac23-x\right)\cdot0,75-2=\frac12\)
=>\(\left(\frac53-x\right)\cdot\frac34=2+\frac12=\frac52\)
=>\(\frac53-x=\frac52:\frac34=\frac52\cdot\frac43=\frac{5\cdot2}{3}=\frac{10}{3}\)
=>\(x=\frac53-\frac{10}{3}=-\frac53\)
e: \(x+75\%\cdot x=-1,6\)
=>\(x+0,75x=-1,6\)
=>1,75x=-1,6
=>\(x=-\frac{160}{175}=-\frac{32}{35}\)
f: \(\left(x-\frac23\right)\left(2x+1\right)=0\)
=>\(\left[\begin{array}{l}x-\frac23=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-\frac12\end{array}\right.\)
g: \(1-\left(2x+\frac12\right)^2=\frac34\)
=>\(\left(2x+\frac12\right)^2=1-\frac34=\frac14\)
=>\(\left[\begin{array}{l}2x+\frac12=\frac12\\ 2x+\frac12=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac12-\frac12=0\\ 2x=-\frac12-\frac12=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\end{array}\right.\)
h: \(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2020}{2021}\)
=>\(1-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2020}{2021}\)
=>\(1-\frac{1}{5x+6}=\frac{2020}{2021}\)
=>\(\frac{1}{5x+6}=\frac{1}{2021}\)
=>5x+6=2021
=>5x=2015
=>x=403