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Công thức tống quát:
\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)
Theo đó, ta có:
\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)
\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)
\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)
\(....................\)
\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)
Nhân lần lượt các đẳng thức trên, ta được:
\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(2A=1-\frac{1}{2n+1}\)
\(A=\frac{1}{2}-\frac{1}{\left(2n+1\right).2}< \frac{1}{2}\)
Vậy:...
- Hok tốt ~
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
=>\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}+\frac{1}{2n+1}\)
=>\(2A=1-\frac{1}{2n-1}\)
=>\(2A=\frac{2n}{2n+1}\)
=>\(A=\frac{2n}{4n+2}=\frac{2n}{2\left(n+1\right)}=\frac{n}{n+1}< \frac{1}{2}\)
zậy A<1/2
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(a_{n-1}=\frac{1}{1+2+3+...+n}=\frac{2}{n\left(n+1\right)}\)=>\(1-a_{n-1}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)........\left(1-\frac{2}{2006.2007}\right)\)
\(=\left(\frac{1.4}{2.3}\right)\left(\frac{2.5}{3.4}\right)\left(\frac{3.6}{4.5}\right)........\left(\frac{2005.2008}{2006.2007}\right)\)\(=\frac{\left(1.2.3......2005\right)\left(4.5.6.....2008\right)}{\left(2.3.4.....2006\right)\left(3.4.5....2007\right)}=\frac{1.2008}{2006.3}=\frac{1004}{3009}\)
\(\left(2x-1\right)\left(3x-2\right)-\left(6x-1\right)\left(x-3\right)=17\)
\(\Rightarrow\left(6x^2-4x-3x+2\right)-\left(6x^2-18x-x+3\right)=17\)
\(\Rightarrow6x^2-4x-3x+2-6x^2+18x+x-3=17\)
\(\Rightarrow\left(6x^2-6x^2\right)+\left(18x-4x-3x+x\right)-\left(3+2\right)=17\)
\(\Rightarrow12x-1=17\)
\(\Rightarrow12x=18\)
\(\Rightarrow x=\frac{3}{2}\)
6x2 - 4x - 3x + 2 - 6x2 - 18x - x + 3 = 17
=>>> -27x + 5 = 17 =>>> -27x = 12 =>>> x = -4/7
Vậy: x = -4/7
=200.6339
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