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\(\frac{1}{1x2}x\frac{4}{2x3}x\frac{9}{3x4}x...x\frac{10000}{100x101}=\frac{1x1}{1x2}x\frac{2x2}{2x3}x\frac{3x3}{3x4}x...x\frac{100x100}{100x101}\)
=\(\frac{1x2x3x...x100}{1x2x3x...x100}x\frac{1x2x3x...x100}{2x3x4x...x101}=1x\frac{1}{101}=\frac{1}{101}\)
\(Giải\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)
\(A=0+0+0+...+0+0\)
\(\Rightarrow A=0\)
\(a.\)\(A< 1\)
b. \(A< \frac{3}{4}\)
a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)
có :
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)
nên :
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< 1-\frac{1}{2011}\)
\(\Rightarrow A< \frac{2010}{2011}< 1\)
b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\)
\(\frac{3}{4}=1-\frac{1}{4}\)
\(\frac{1}{4}>\frac{1}{2011}\)
nên :
\(A>\frac{3}{4}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=\(1-\frac{1}{9}\)
=\(\frac{8}{9}\)
OK XONG NHỚ CHO MIK NHA
\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}\)
=\(\frac{8}{9}\)
= 29 nhé
Mình ko chắc nhé , nên mình sai đừng k mình sai !
\(\frac{1}{3×4}+\frac{1}{4×5}+...+\frac{1}{x+\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{10}{30}-\frac{9}{30}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1\)
\(x=29\)
Ở đây đề bị lỗi là : 1/x+(x+1) đáng lẽ phải là 1/x.(x+1) thì mới đúng .
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)
\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
1.
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{2012}{2013}\)
\(A=\frac{1.2.3.4.....2012}{2.3.4.5......2013}\)
\(A=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(B=\frac{2012\left(2013-2012\right)}{2012\left(2011+2\right)}\)
\(B=\frac{2013-2012}{2011+2}\)
\(B=\frac{1}{2013}\)
\(Vì:\frac{ 1}{2013}=\frac{1}{2013}\)
\(\Rightarrow\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}......\frac{2012.2013}{2013.2013}=\frac{2012.2013-2012.2012}{2012.2011+2012.2}\)
\(Hay: A=B\)
\(A=\frac{1\times2}{2\times2}\times\frac{2\times3}{3\times3}\times\frac{3\times4}{4\times4}\times\frac{4\times5}{5\times5}\times...\times\frac{2012\times2013}{2013\times2013}\)
\(\Rightarrow A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2012}{2013}\)
\(\Rightarrow A=\frac{1\times2\times3\times4\times...\times2012}{2\times3\times4\times5\times...\times2013}\)
\(\Rightarrow A=\frac{1}{2013}\)
\(B=\frac{2012\times2013-2012\times2012}{2012\times2011+2012\times2}\)
\(\Rightarrow B=\frac{2012\times\left(2013-2012\right)}{2012\times\left(2011+2\right)}\)
\(\Rightarrow B=\frac{2012\times1}{2012\times2013}\)
\(\Rightarrow B=\frac{1}{2013}\)