Tìm x biết
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
PS : Không cần làm đẹp, càng nhanh càng tốt, quan trọng là đúng!
Tìm số tự nhiên x biết \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)
\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)
=>x+1=2005
=>x=2004
1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015
Đ/A là 2004
chúc đồng chí Chế Minh Hải học tốt
TìM X : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....................+\frac{2}{x.\left(x-1\right)}=1\frac{2003}{2005}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{20}.\left(1+2+3+...+20\right)\\\)
Tính càng nhanh càng tốt
nhanh và đúng nhất mình k 2 lần~~~~
Từ công thức:\(1+2+........+n=\frac{n.\left(n+1\right)}{2}\)
Cho \(n\in\)N*.CMR:\(\frac{1}{n}.\left(1+2+...+n\right)=\frac{n+1}{2}\)
Ta có:\(\frac{1}{n}.\left(1+2+......+n\right)=\frac{1}{n}.\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\)
Ta có:\(1+\frac{1}{2}\left(1+2\right)+......+\frac{1}{20}.\left(1+2+.....+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+........+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=1+\frac{3}{2}+...............+\frac{21}{2}\)
\(=\frac{2+3+......+21}{2}\)
\(=\frac{230}{2}=165\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)
\(\Leftrightarrow4006x+4006=4010x-4010\)
\(\Leftrightarrow-4x=-8016\)
\(\Leftrightarrow x=2004\)
Vậy x = 2004
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)
\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)
\(\left(x+1\right).7=4020\)
\(\Rightarrow x=....\)
Tìm \(x\)biết:
\(a,\)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2005}{2007}\)
\(b,\)\(\left[\frac{6:\frac{3}{5}-1\frac{1}{16}.\frac{3}{7}}{4.\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}}-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{5}\right).\frac{12}{49}}{3\frac{1}{3}+\frac{2}{8}}\right].x=2\frac{23}{96}\)
Bạn nào làm nhanh và đầy đủ nhất thì mình tik cho
giúp tớ giải bài này nha càng nhanh càng tốt !
Tính:S=\(\left(1-\frac{1}{^{2^2}}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{2016^2}\right)\)
Mình cần gấp các bn ạ !
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
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
Tìm giá trị nguyên của x biết :
\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
Mình cần gấp
Bnạ nào làm nhanh và đúng mình sẽ tick
Theo đề ra ,ta có :
- 1 / 12 < x < 1 / 8 mà x có giá trị nguyên
=> x = 0
Tìm x
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+.....+\(\frac{2}{x\left[x+1\right]}\)=\(\frac{2003}{2005}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)