1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+...+2012
S=1/2*2+1/3*3+1/4*4+.....+1/2012*2012
Bài 1
a, Tính P=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+....+1/2012(1+2+3+...+2012)
b,Tìm x thỏa mãn 4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=2^x
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
A=[2012+2011/2+2010/3+2009/4+...1/2012]:[1/2+1/3+1/4+...+1/2012+1/2013]
HOI A CHIA 3 DU BAO NHIEU ?
Tìm x:
x . (1/2+1/3+1/4+. . .+1/2011+1/2012)
2012/1+2011/2+2010/3+2009/4+ . . . +2/2011+1/2012
=1
(1-1/1+2)(1-1/1+2+3)(1-1/1+2+3+4)+.......+(1-1/1+2+3+....+2012)
\(\dfrac{1-1}{1+2}+\dfrac{1-1}{1+2+3}+\dfrac{1-1}{1+2+3+4}+...+\dfrac{1-1}{1+2+3+...+2012}\)
\(=\dfrac{0}{1+2}+\dfrac{0}{1+2+3}+\dfrac{0}{1+2+3+4}+...+\dfrac{0}{1+2+3+4+...+2012}\)
\(=0+0+0+...+0\)
\(=0\)
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Bài này dễ mà bạn. Cơ mà hình như bạn ghi sai đề, sao khúc đầu thì nhân mà khúc cuối lại cộng thế?
Tính \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2013\sqrt{2012}+2012\sqrt{2013}}\)
= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)
= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)
= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)
= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)
= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)
Tìm x biết: (1/2+1/3+1/4+...+1/2014).x =2013/1+2012/2+2011/3+...+2/2012+1/2013
trước tiên bạn phải tính:
2013/1+2012/2+2011/3+.....+2/2012+1/2013
=1+2012/2)+(1+2011/3)+.....+(1+2/2012)+(1+1/2013) +1 {BƯỚC NÀY TÁCH 2013 RA LÀM 2013SỐ1 ĐỂ CÔNG VS CÁC THỪA SỐ CÒN LẠI}
=2014/2+2014/3+...+2014/2012+2014/2013+2014/2014
=2014.(1/2+1/3+....+1/2012+1/20131/2014
suy ra x=2014
Rút gọn A= (1/2+ 1/3+ 1/4+.....+ 1/2013)/ ( 2012+ 2012/2 + 2011/ 3+....+ 1/ 2013)
Rút gọn A= (1/2+1/3+1/4+...+1/2013)/(2012+2012/2+2011/3+...+1/2013)