Tìm x , biết:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+......+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{4949}{19800}\)
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\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
thay A vào ta được
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45
=>(1/1.2-1/2.3+1/2.3-1/3.4+....+1/8.9-1/9.10).x=22/45
=>(1/1.2-1/9.10).x=22/45
=>22/45.x=44/45
=>x=2
4+2^2+2^3+....+2^20=2^n
=>2^2+2^2+2^3+....+2^20=2^n
đặt 2^2+2^3+....+2^20
=>2A-A=2^21-2^2
khi đó A=2^2+2^21-2^2=2^21=2^n
=>n=21
Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)
=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)
2B = \(\frac{22}{45}\)
B = \(\frac{22}{45}:2\)
=> B = \(\frac{11}{45}\)
Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)
=> x = \(\frac{22}{45}:\frac{11}{45}\)
=> x = \(\frac{2}{1}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)
Vậy x=2
a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0
=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0
b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4
=1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)
= (x-1)x(x+1)(x+2)
=> A=x(x+1)(x+2)(x-1)/4