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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 = \(\dfrac{502}{3}\)
a)13x3x32,27+67,63x39
=39x32,27+67,63x39
=39x(32,27+67,63)
=39x100
=3900
b,= 1- [ 1/2 x 1/3 x1/4 x..... x 1/100 ]
=1/2 x 2/3 x 3/4 x .......x 99/100
= 1x2x3x......x99 / 2x3x4x...... x100 [ rút gọn ]
= 1/100
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
\(71+52,5\times4=\frac{x+140}{x}+210\)
\(71+210=\frac{x+140}{x}+210\)
\(=>\frac{x+140}{x}=71\)
\(71=\frac{142}{2}\)\(\Rightarrow x=142-140=2\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\)\(\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(15+5x=50\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\Rightarrow1+x+2+x+3+x+4+x+5+x=50\)
\(\Rightarrow\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(\Rightarrow15+5x=50\)
\(\Rightarrow5x=50-15\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=35:5\)
\(\Rightarrow x=7\).
X x\(\frac{1}{2}\)+ X x\(\frac{3}{4}\)=\(\frac{6}{7}\)
=>X x(\(\frac{1}{2}\)+ \(\frac{3}{4}\)) =\(\frac{6}{7}\)
=>X x \(\frac{5}{4}\)=\(\frac{6}{7}\)
=>X=\(\frac{6}{7}\): \(\frac{5}{4}\)
=>X=\(\frac{24}{35}\)
\(x+0,9=20,17\Leftrightarrow x=20,17-0,9=19,24\)
\(47,5-x=4,64\Leftrightarrow x=47,5-4,64=42,86\)
\(3x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(a.x=20,17-0,9=19,27\)
\(b.47,5-x=4,56\Leftrightarrow x=42,86\)
\(c.3.x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)