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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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
=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)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
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
Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)
Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)
\(\left(y+\frac{3}{7}\right)^{468}=0\)
=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)
\(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)
=> \(x-\frac{2}{5}=0\)
\(y-\frac{3}{7}=0\)
=> \(x=\frac{2}{5}\)
\(y=\frac{3}{7}\)
Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)
\(\frac{34-x}{30}=\frac{25}{30}\)
34 - x = 25
x = 34 - 25 = 9
\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)
\(\frac{x+13}{34}=\frac{24}{34}\)
x + 13 = 24
x = 24 - 13 = 11
\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)
\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(2A=1-\frac{1}{81}=\frac{80}{81}\)
\(A=\frac{80}{81}\div2=\frac{40}{81}\)
\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)
\(4x=\frac{56}{81}-\frac{40}{81}\)
\(4x=\frac{16}{81}\)
\(x=\frac{16}{81}\div4=\frac{4}{81}\)
a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
=> (2,85 - x - 0,5) : 0,25 = 60
=> (2,85 - 0,5) - x = 60 . 0,25
=> 2,35 - x = 15
=> x = 2,35 - 15
=> x = -12,65
Vậy x = -12,65
b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)
\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)
\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)
\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)
\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)
\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)
\(\Rightarrow x=\frac{26}{6}\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}\)
c) \(35\left(2\frac{1}{5}-x\right)=32\)
\(\Rightarrow2\frac{1}{5}-x=32\div35\)
\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)
\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)
\(\Rightarrow x=\frac{9}{7}\)
Vậy \(x=\frac{9}{7}\)
d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)
\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)
\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)
\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)
\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)
\(\Rightarrow x=\frac{56}{405}\)
Vậy \(x=\frac{56}{405}\)
e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)
\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)
\(\Rightarrow4.\frac{12}{11}=11-5\div x\)
\(\Rightarrow11-5\div x=\frac{48}{11}\)
\(\Rightarrow5\div x=11-\frac{48}{11}\)
\(\Rightarrow5\div x=\frac{73}{11}\)
\(\Rightarrow x=5\div\frac{73}{11}\)
\(\Rightarrow x=\frac{55}{73}\)
Vậy \(x=\frac{55}{73}\)
a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125
(2,85 - x - 0,5) : 0,25 = 60
(2,85 - x - 0,5) = 60 x 0,25
(2,85 - x - 0,5) = 15
2,35 - x = 15
x = 2,35 - 15
x = -12,65
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$