Tìm x thuộc Z biết:
2/2x3+2/3x4+2/4x5+...+2/x(x+1) = 2008/2010
K
Khách
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LL
1
22 tháng 7 2016
\(P=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
NB
2
5 tháng 4 2016
=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100
1/2-1/a+1=49/100
1/a+1 = 1/2-49/100
1/a+1=1/100
a+1=100
a=99
5 tháng 4 2016
=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100
1/2-1/a+1=49/100
1/a+1 = 1/2-49/100
1/a+1=1/100
a+1=100
a=99
CH
1
BD
30 tháng 5 2023
\(\left(x+\dfrac{1}{2\times3}\right)+\left(x+\dfrac{1}{3\times4}\right)+\left(x+\dfrac{1}{4\times5}\right)+\left(x+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x+\dfrac{1}{2\times3}+x+\dfrac{1}{3\times4}+x+\dfrac{1}{4\times5}+x+\dfrac{1}{5\times6}=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
\(x\times4+\dfrac{4}{12}=\dfrac{25}{3}\)
\(x\times4=\dfrac{25}{3}-\dfrac{4}{12}\)
\(x\times4=\dfrac{25}{3}-\dfrac{1}{3}\)
\(x\times4=\dfrac{24}{3}\)
\(x\times4=8\)
\(x=8\div4\)
\(x=2\)
:))
PA
8 tháng 7 2016
1.
a.
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
b.
Tích có 100 thừa số
=> n = 100
\(\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times\left(100-100\right)\)
\(=\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times0\)
\(=0\)
2.
a.
\(135\times789789-789\times135135=1001\times\left(135\times789-789\times135\right)=1001\times0=0\)
b.
\(\left(28\times9696-96\times2828\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left[101\times\left(28\times96-96\times28\right)\right]\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left(101\times0\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\)
3.
a.
\(\left[\left(x+32\right)-17\right]\times2=42\)
\(\left(x+32\right)-17=\frac{42}{2}\)
\(\left(x+32\right)-17=21\)
\(x+32=21+17\)
\(x+32=38\)
\(x=38-32\)
\(x=6\)
b.
\(125+\left(145-x\right)=175\)
\(145-x=175-125\)
\(145-x=50\)
\(x=145-50\)
\(x=95\)
25 tháng 7 2015
a. Vì
1/2<2/3
3/4<4/5
.........
99/100<100/101 nên M<N
b.M.N=\(\frac{1.2.3.4......100}{2.3.4.5......101}\)=\(\frac{1}{101}\)
4 tháng 2 2019
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2008}{2010}.\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
=> x + 1 = 2010
=> x = 2009
4 tháng 2 2019
Ta có : \(\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{x\times\left(x+1\right)}=\frac{2008}{2010}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{1004}{1005}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}=\frac{1}{2010}\)
\(\Rightarrow x+1=2010\)
\(\Rightarrow x=2010-1=2009\)
\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\\ 2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\\ \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2010}:2\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1004}{2010}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1004}{2010}\\ \dfrac{1}{x+1}=\dfrac{1}{2010}\\ \Rightarrow x+1=2010\\ \Rightarrow x=2009\)
nhìn đề bài ko hỉu j hết