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ta đặt: A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/2005.2006.2007
2.A = 2(1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/2005.2006.2007)
2.A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/2005.2006.2007
= (1/1.2 - 1/2.3) + (1/2.3 - 1/3.4) +...+ (1/2005.2006- 1/2006.2007)
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... +1/2005.2006 - 1/2006.2007
= 1/1.2 - 1/2006.2007
=> A = (1/1.2 - 1/2006.2007):2
A = 1/4 - 1/1003.2007
Đặt B = 1/1.2 + 1/2.3+ 1/ 3.4 ..... + 1/2006.2007
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/2006-1/2007)
=1/1-1/2+1/2-1/3+1/3-1/4+....+1/2006-1/2007
=1/1-1/2007
= 2006/2007
thay vào phương trình ta có phương trình trở thành:
(1/4 - 1/1003.2007).x = 2006/2007
..........
còn lại bạn tính nhé
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{2005.2006.2007}\)
\(B=1.2+2.3+3.4+....+2006.2007\)
Ta có : \(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2006.2007}\right)\)
\(B=1.2+2.3+3.4+....+2006.2007\)
\(=\frac{1.2.3+2.3.\left(4-1\right)+3.5.\left(5-2\right)+...+2006.2007.\left(2008-2005\right)}{3}\)
\(=\frac{1.2.3+2.3.4-1.2.3+3.4.5-...+2006.2007.2008-2005.2006.2007}{3}\)
\(=\frac{2006.2007.2008}{3}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2006.2007}\right)x=\frac{2006.2007.2008}{3}\)
\(\Rightarrow x=\frac{2006.2007.2008}{3}:\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2006.2007}\right)\right]\)(tự tính)
Đặt \(NCTK=VT\)
\(\Rightarrow2NCTK=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...\)
\(+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(\Rightarrow2NCTK=\frac{1}{2}-\)\(\frac{1}{2006.2007}\)
\(\Rightarrow NCTK=\frac{1}{4}-\frac{1}{2.2006.2007}\)
Đặt \(KN=1.2+2.3+...+2006.2007\)
\(3KN=1.2.3+2.3.\left(4-1\right)+...+2006.2007\left(2008-2005\right)\)
\(=2006.2007.2008\)
\(KN=\frac{2006.2007.2008}{3}\)
...
\(\frac{2}{1.2}+\frac{2}{2.3}+..........+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{x\left(x+1\right)}\right)=\frac{4028}{2015}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{x}-\frac{1}{x+1}=\frac{4028}{2015}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\Rightarrow x=2014\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}\right)=1\frac{2013}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{2013}{2015}\div2\)
\(1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
\(x+1=2015\)
\(x=2015-1\)
\(x=2014\)
Ta có:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); ...; \(\frac{2}{2005.2006.2007}=\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2006.2007}\right)\)
\(A=\frac{1}{2}\left(\frac{1003.2007-1}{2006.2007}\right)\)
B=1.2+2.3+3.4+...+2006.2007=\(\frac{2006.2007.2008}{3}\)
Ta có: A.x=B => x=B:A = \(\frac{2006.2007.2008}{3}:\left\{\frac{1}{2}.\frac{1003.2007-1}{2006.2007}\right\}=\frac{2006.2007.2008}{3}.\frac{2.2006.2007}{1003.2007-1}\)
=> \(x=\frac{2.2006^2.2007^2.2008}{6039060}=2676.2007^2\)
ta đặt: A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/2005.2006.2007
2.A = 2(1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/2005.2006.2007)
2.A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/2005.2006.2007
= (1/1.2 - 1/2.3) + (1/2.3 - 1/3.4) +...+ (1/2005.2006- 1/2006.2007)
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... +1/2005.2006 - 1/2006.2007
= 1/1.2 - 1/2006.2007
=> A = (1/1.2 - 1/2006.2007):2
A = 1/4 - 1/1003.2007
Đặt B = 1/1.2 + 1/2.3+ 1/ 3.4 ..... + 1/2006.2007
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/2006-1/2007)
=1/1-1/2+1/2-1/3+1/3-1/4+....+1/2006-1/2007
=1/1-1/2007 = 2006/2007
thay vào ta được phương trình trở thành:
(1/4 - 1/1003.2007).x = 2006/2007
..........
Bài này không tính nhé tth nghĩ nát óc mới ra :3
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2005.2006.2007}\right)x=1.2\left(3-0\right)+2.3\left(4-1\right)+...+2006+2007\left(2008-2005\right)\)\(3\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2005.2006.2007}\right)x=2\left(1.2\left(3-0\right)+2.3+...+2006+2007\right)\)
\(2\left(1.2.3+2.3.4-1.2.3+...+2006+2007.2008-2005.2006.2007\right)\)
Đến đây rồi tự làm tiếp đi nhé
Đặt \(A=1.2+2.3+3.4+......+98.99\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+......+98.99.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+98.99.\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100=970200\)
\(\Rightarrow A=\frac{970200}{3}=323400\)
Ta có: \(\frac{\left(1.2+2.3+3.4+....+98.99\right).x}{323400}=323400\)
\(\Leftrightarrow\frac{323400.x}{323400}=323400\)\(\Leftrightarrow x=323400\)
Vậy tập nghiệm của phương trình là \(S=\left\{323400\right\}\)
Gọi \(A=1.2+2.3+3.4+...+98.99\)
\(3A=1.2.3+2.3.3+3.4.3+...+98.99.3\)
\(=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)
\(=1.2.3-1.2.0+2.3.4-2.3.1+3.4.5-3.4.2+...+98.99.100-98.99.97\)
\(=98.99.100-1.2.0\)
\(=970200\)
\(A=\frac{970200}{3}\)
\(=323400\)
Ta có : \(\frac{A.x}{323400}=323400\)
\(\Leftrightarrow\frac{323400.x}{323400}=323400\)
\(\Leftrightarrow x=323400\)
Vậy phương trình trên có 1 nghiệm \(x=323400\)