tìm x tuộc N :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
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8 tháng 6 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(-\frac{1}{x+1}=\frac{2001}{4006}-\frac{1}{2}\)
\(-\frac{1}{x+1}=-\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(\Rightarrow x=2012\)
NM
8 tháng 6 2016
Ta có: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}:2\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{2003}{4006}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2003}{4006}-\frac{2001}{4006}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2002
Vậy x = 2002
Duyệt nha !!!
chúc hk tốt!!!
8 tháng 6 2016
13 +16 +110 +....+1x(x+1):2 =20012003
26 +212 +220 +....+2x(x+1) =20012003
2(12.3 +13.4 +14.5 +....+1x(x+1) )=20012003
12 −13 +13 −14 +14 −15 +....+1x −1x+1 =20012003 :2=20014006
12 −1x+1 =20014006
1x+1 =12 −20014006 =12003
=> x+1 = 2003
=> x = 2003 - 1
=> x = 2002
Xin 1 tích đúng
8 tháng 6 2016
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2001}{2003}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2001}{2003}\)
\(\Rightarrow2x=4004\)
\(\Rightarrow x=2002\)
23 tháng 4 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
26 tháng 5 2016
(x+4/2000 + 1)+(x+3/2001 + 1) = (x+2/2002 + 1)+(x+1/2003)+1
(x+2004/2000) + (x+2004/2001) = (x+2004/2002) + (x+2004/2003)
(x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
+ Với x+2004=0 suy ra x=-2004. Ta có 0.(1/2000+1/2001)=0.(1/2002+1/2003), đúng
+ Với x+2004 khác 0 thì (x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
1/2000+1/2001 = 1/2002+1/2003, vô lí vì 1/2000+1/2001 > 1/2002+1/2003
Vậy x=-2004
TN
19 tháng 6 2017
Áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rút gọn rồi quy đồng làm nốt
26 tháng 5 2016
\(PT\Leftrightarrow\frac{x+4+2000}{2000}+\frac{x+3+2001}{2001}=\frac{x+2+2002}{2002}+\frac{x+1+2003}{2003}\)
<=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004.
26 tháng 5 2016
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)
\(\Rightarrow x=-2004\)
16 tháng 5 2016
Đặt A=1/3+1/6+1/10+...+2/x*(x+1)
1/2A=1/3*2+1/6*2+1/10*2+...+2/2*x*(x+1)
1/2A=1/6+1/12+1/20+...+1/x*(x+1)
1/2A=1/2*3+1/3*4+1/4*5+...+1/x*(x+1)
1/2A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)
1/2A=1/2-1/x+1
A=(1/2-1/x+1):1/2
A=1-2/x+1
Ta có A=1999/2001
Hay 1-2/x+1=1999/2001
2/x+1=1-1999/2001
2/x+1=2/2001
=>x+1=2001
=>x=2000
HT
16 tháng 5 2016
Cho A = 1/3+1/6+1/10+...+2/x(x+1)
1/2A= 1/3.2+1/6.2+1/10.2+...+2/x(x+1)2
1/2A= 1/6+1/12+1/20+...+1/x(x+1)
1/2A= 1/2.3+1/3.4+1/4.5+...+1/x(x+1)
1/2A= 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1
1/2A= 1/2-1/x+1
A = (1/2-1/x+1)/1/2
A = 1-2/x+1
Mà A=1999/2001
=> 1-2/x+1= 1999/2001
2/x+1= 1-1999/2001
2/x+1= 2/2001
=>x+1=2001
=>x = 2000
27 tháng 2 2020
a, Ta có : \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
=> \(\frac{4\left(x+1\right)}{12}+\frac{9\left(2x+1\right)}{12}=\frac{2\left(2x+3\left(x+1\right)\right)}{12}+\frac{7+12x}{12}\)
=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3\left(x+1\right)\right)+7+12x\)
=> \(4\left(x+1\right)+9\left(2x+1\right)=2\left(2x+3x+3\right)+7+12x\)
=> \(4x+4+18x+9=4x+6x+6+7+12x\)
=> \(4x+18x-12x-6x-4x=6+7-4-9\)
=> \(0x=0\) ( Luôn đúng với mọi x )
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
=> \(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1-\frac{x}{2003}+1\)
=> \(\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\frac{-x}{2003}+1\)
=> \(\frac{2-x}{2001}+\frac{2001}{2001}=\frac{1-x}{2002}+\frac{2002}{2002}+\frac{-x}{2003}+\frac{2003}{2003}\)
=> \(\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
=> \(\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)
=> \(\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
=> \(2003-x=0\)
=> \(x=2003\)
Vậy phương trình có tập nghiệm là \(S=\left\{2003\right\}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
<=>\(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}\cdot\frac{1}{2}=\frac{2001}{4006}\)
<=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
<=>\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
<=>\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
<=>\(\frac{1}{x+1}=\frac{1}{2003}\)
<=>x+1=2003
<=>x=2002