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=\(\frac{6\left(1+8+27+64\right)}{12\left(1+16+54+128\right)}\)
=\(\frac{6.100}{12.199}\)
=\(\frac{50}{199}\)
Tk mình với nha mọi người!!!!!
\(\frac{1x2x3+2x4x6+3x6x9+4x8x12}{1x3x4+4x6x8+6x9x12+8x12x16}\)
\(\frac{6x\left(1+8+27+64\right)}{12x\left(1+16+54+128\right)}=\frac{6x100}{12x199}=\frac{50}{199}\)
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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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
\(=\frac{6}{5}\times\frac{7}{6}\times...\times\frac{11}{10}\)(lại lỗi đề)
\(=\frac{6×7×...×11}{5×6×...×10}\)
\(=\frac{11}{5}\)
\(1\frac{1}{5}\cdot1\frac{1}{6}\cdot1\frac{1}{7}\cdot1\frac{1}{8}\cdot1\frac{1}{9}\cdot1\frac{1}{10}\)
\(=\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}\cdot\frac{9}{8}\cdot\frac{10}{9}\cdot\frac{11}{10}\)
\(=\frac{6\cdot7\cdot8\cdot9\cdot10\cdot11}{5\cdot6\cdot7\cdot8\cdot9\cdot10}\)
\(=\frac{11}{5}\)
\(=\dfrac{1}{2x1x3x2}+\dfrac{1}{2x2x3x3}+\dfrac{1}{2x3x3x4}+...+\dfrac{1}{2x18x3x19}+\dfrac{1}{2x19x3x20}=\)
\(=\dfrac{1}{2x3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{18x19}+\dfrac{1}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{20-19}{19x20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)=\)
\(=\dfrac{1}{6}x\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}x\dfrac{19}{20}=\dfrac{19}{120}\)