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=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(Giải\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)
\(A=0+0+0+...+0+0\)
\(\Rightarrow A=0\)
\(a.\)\(A< 1\)
b. \(A< \frac{3}{4}\)
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A>1-\frac{1}{2015}\)
Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)
\(\frac{1}{4}\)+ \(\frac{1}{9}\)+ \(\frac{1}{16}\)+...+ \(\frac{1}{2401}\)+ \(\frac{1}{2500}\)
Dãy số trên có :
50 - 2 + 1 = 49 số hạng
Tổng các tử số của sô hạng trên là :
1 x 49 = 49
Mà 49 < 2401; (2401 là mấu cố số hạng kế cuối cùng) mà 2401 : 49 = 49
Kết luận tổng dãy số trên có tử số < mẫu số -> tổng dãy số bé hơn 1
Dấu cần điền "<"
=1/2 -1/2 +1/3 -1/3 +....+1/50 -1/50=0
0<1
suy ra 1/2*2 +1/3*3 +.....+1/49*49 +1/50*1/50 <1
không chắc lắm nhưng nếu muốn bạ có thể tính "tổng xích ma" trên máy tích cầm tay casio fx 720
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
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 :
\(A>\frac{3}{4}\)
\(C=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{100\times100}\\ C< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\\ C< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ C< 1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)