0,12

0,05

0,306

TL:

\(\frac{12}{100}\)= 0,12

\(\frac{5}{100}\)= 0,05

\(\frac{306}{1000}\)= 0,306

-HT-

BẠN LẤY CÁI 5/30+45/270 NHA

NHỚ K NHA

=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 = \(\dfrac{502}{3}\)

a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)

   A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)

   A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)

   A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)

  A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)

 A = 2

a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b)  \(\frac{99}{100}\)< \(\frac{100}{99}\);   c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\); 

d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)

e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)

3/4 × 8/9 × 15/16 × ... × 99/100

= 1×3/2×2 × 2×4/3×3 × 3×5/4×4 × ... × 9×11/19×10

= 1×2×3×...×9/2×3×4×...×10 × 3×4×5×...×11/2×3×4×...×10

= 1/10 × 11/2

= 11/20

bạn làm rõ giúp mình nhé

\(\frac{1}{10}=\frac{10}{100}\) ; \(\frac{2}{10}=\frac{20}{100}\)

Trong các đấp án trên chỉ có a, là hợp lý vì \(\frac{10}{100}<\frac{15}{100}<\frac{20}{200}\)

Giair:

\(\frac{1}{10}=0,1\)

\(\frac{2}{10}=0,2\)

Ta có 0,1 <0,15<0,2 Hoặc 1/10 <15/10<2/10

Vậy số ở giữa là 15/10

Học tốt

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: 

\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)

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