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Mk cg tính ra kết quả này nhg thấy sai sai nên cg chưa đăng nè
\(\Rightarrow2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2x-2\right).2x}\right)=\frac{1}{8}.2\).2
\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+...\frac{2}{\left(2x-2\right).2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\Rightarrow\frac{1}{2x}=\frac{1}{2.2}\)
\(\Rightarrow x=2\)
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
Gọi biều thức trên là A, ta có:
A=(1/2.4+1/4.6+1/6.8+1/8.10+1/10.12)x=2
2A=(2/2.4+2/4.6+2/6.8+2/8.10+2/10.12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2
2A=(1/2-1/12)x=2
2A=5/12x=2
=>A=5/24x=1
=>x=1:5/24=24/5
=>1/2.(5/12).x=1
5/24.x=1
x=1:5/24
x=24/5
lưu ý, 1/2.5/12 là tính xong phần 1/2.4 +...+1/10.12 rùi nhé
=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)
=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
=>\(\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
=>\(\frac{1}{2x}=\frac{1}{4}\)
=> \(2x=4\)
=> \(x=2\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018
1/6+1/12+1/20+...+1/x(x+1)=504/1009
1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009
1/2-1/x+1=504/1009
x-1/2(x+1)=504/1009
-> 1009(x-1)=504.2(x+1)
1009x-1009=1008x+1008
1009x-1008x=1008+1009
->x=2017
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)
mình đang cần gấp
\(\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{x\left(x+2\right)}=\frac{1}{10}\left(ĐKXĐ:x\ne0;x\ne-2\right)\)
\(\Rightarrow\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{2}{10}\).
\(\Rightarrow\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{1}{5}\).
\(\Rightarrow\frac{1}{4}-\frac{1}{x+2}=\frac{1}{5}\).
\(\Rightarrow\frac{5\left(x+2\right)}{20\left(x+2\right)}-\frac{20}{20\left(x+2\right)}=\frac{4\left(x+2\right)}{20\left(x+2\right)}\)..
\(\Rightarrow5\left(x+2\right)-20=4\left(x+2\right)\).
\(\Rightarrow5x+10-20=4x+8\).
\(\Rightarrow5x-4x=8-10+20\).
\(\Rightarrow x=18\)(thỏa mãn ĐKXĐ) (ĐKXĐ : Điều kiện xác định).
Vậy \(x=18\).