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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 = \(\dfrac{502}{3}\)
\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)
a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(\frac{12}{15}\times\frac{22}{47}+\frac{25}{47}:\frac{15}{12}\)
\(\frac{12}{15}\times\frac{22}{47}+\frac{25}{47}\times\frac{12}{15}\)
\(=\frac{12}{15}\times\left(\frac{22}{47}+\frac{25}{47}\right)\)
\(=\frac{12}{15}\times\frac{22+25}{47}\)
\(=\frac{12}{15}\times1=\frac{12}{15}\)
TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
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) \(\left(15-x\right)+\left(x-12\right)=7-\left(-8+x\right)\)
\(15-x+x-12=7+8-x\)
\(x=12\)
b) \(x-\left\{55-\left[49+\left(-28-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
\(x-55+49-28-x=13-47-25+32-x\)
\(x=7\)
\(a,\left(15-x\right)+\left(x-12\right)=7-\left(-8+x\right)\)
\(15-x+x-12=7+8-x\)
\(15-x=3\)
\(x=12\)
\(b,x-\left\{55-\left[49+\left(-28-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
\(x-\left[55-\left(49-28-x\right)\right]=13-\left[47+\left(25-32+x\right)\right]\)
\(x-\left(55-49+28+x\right)=13-\left(47+25-32+x\right)\)
\(x-\left(34+x\right)=13-\left(40+x\right)\)
\(x-34-x=13-40-x\)
\(-27-x=-34\)
\(x=-27--34\)
\(x=7\)
\(c,\left|x-10\right|-\left|-12\right|=4\)
\(\left|x-10\right|-12=4\)
\(\left|x-10\right|=16\)
\(\Rightarrow x-10=16\)hoặc \(x-10=-16\)
\(x=26\)hoặc \(x=-6\)
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
cái này là toán đó nha mik ghi nhầm
12/15. 22/47+25/47:15/12
=12/15.22/47+25/47.12/15
=12/15.(22/47+25/47)
=12/15.1
=12/15
(. nghĩa là nhân)