Tìm số tự nhiên x biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{2}{x.\left(x+1\right)}\)
Tìm số tự nhiên x biết rằng:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Đặt: A= 1/3 +1/6+1/10+…+2/x(x+1)
A x 1/2 = 1/2.3 + 1/3.4 + 1/4.5 +…+1/x(x+1)
A x1/2 = 1/2-1/3+1/3-1/4+1/4-1/5+…..+1/x-1/(x+1)
A x 1/2 = 1/2 – 1/(x+1)
A = (1/2 -1/x+1) : 1/2
A = 1 – 2/(x+1)
Như vậy ta có: 1-2/(x+1) = 1999/2001
Hay: 2/(x+1) = 1-1999/2001
2/(x+1) = 2/2001
Vậy x = 2000
Tích tớ nha!! Cáchgiải chính xác 100%
1/3+1/6+1/10+...+1/x(x+1)=1999/2001
1/2.[1/3+1/6+1/10+...+1/x(x+1)].2=1999/2001
[1/6+1/12+1/20+...+1/x(x+1)].2=1999/2001
[1/2.3+1/3.4+1/4.5+...+1/x.(x+1)].2=1999/2001
[1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1].2=1999/2001
(1/2-1/x+1).2=1999/2001
1/2-1/x+1=1999/2001:2=1999/2001.1/2=1999/4002
1/x+1=1/2-1999/4002
1/x+1=2001/4002-1999/4002==2/4002=1/2001
=>x+1=2001
=>x=2001-1=2000
Vậy x=2000
Tìm x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2010
Tìm số tự nhiên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Giải đầy đủ nha, ko sai đề đâu
tìm số nguyên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
Chưa chắc là đề sai!!!!!!!!!!!!!!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{2003}.\frac{1}{2}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(x+1=2003\)
\(x=2002\)
Tìm x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{X\left(X+1\right)}=\frac{1999}{2001}\)
Tu de bai ta co
1/6+1/12+1/20+...+1/(x*(X+1))=1999/4002
Suy ra 1/(2*3)+1/(3*4)+1/(4*5)+...+1/(x*(x+1))=1999/4002
Suy ra 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=1999/4002
Suy ra 1/2-1/(x+1)=1999/4002
Suy ra 1/(x+1)=1/2001
Suy ra x+1=2001
Suy ra x=2000
TÌm x, biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+ ..... +\(\frac{1}{X.\left(X+1\right)}\)=\(\frac{1999}{2001}\)
\(\frac{2}{2.3}\)+\(\frac{2}{2.6}\)+\(\frac{2}{2.10}\)+ ...... + \(\frac{1}{X.\left(X+1\right)}\)=\(\frac{1999}{2001}\)
\(\frac{2}{2.3}\)\(+\)\(\frac{2}{3.4}\)\(+\) \(\frac{2}{4.5}+...\) \(+\) \(\frac{1}{x\left(x+1\right)}\)=\(\frac{1999}{2001}\)
\(2\)\(.\)(\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\) ....) \(+\)\(\frac{1}{x\left(x+1\right)}\)\(=\)\(\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\)\(\frac{1999}{2001}:2\)
\(\frac{1}{2}-\frac{1}{x+1}\)\(=\frac{1999}{2001}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}\)
\(\frac{1}{x+1}=\frac{2}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
\(\Rightarrow x+1=2001\)
\(\Rightarrow x=2000\)
chúc bạn học giỏi. đúng thì k cho mình nha
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
Tìm x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{199}{2001}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\Leftrightarrow\frac{1}{x+1}=\frac{1}{2001}\)
\(\Leftrightarrow x+1=2001\Rightarrow x=2000\)
Tìm số tự nhiên x biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\times\left(x+1\right):2}=\frac{2001}{2003}\)
Giải chi tiết nha!!!!Ai nhanh mik tick liền!!!!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Tìm x.
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{1999}{2001}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2000