\(\dfrac{1-1}{1+2}+\dfrac{1-1}{1+2+3}+\dfrac{1-1}{1+2+3+4}+...+\dfrac{1-1}{1+2+3+...+2012}\)
\(=\dfrac{0}{1+2}+\dfrac{0}{1+2+3}+\dfrac{0}{1+2+3+4}+...+\dfrac{0}{1+2+3+4+...+2012}\)
\(=0+0+0+...+0\)
\(=0\)
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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?
Tìm x biết: (1/2+1/3+1/4+...+1/2014).x =2013/1+2012/2+2011/3+...+2/2012+1/2013
Tính giá trị biểu thức:
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{2}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(H=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)
\(I=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2011}}\)
Help me!
Rút gọn A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2013}}{2012+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}}\)
Cho \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2012}}\)
CMR A < 1
Tính A = ((1/2^2)-1).((1/3^2)-1)...((1/2012^2)-1)
So sánh :
\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)và \(\dfrac{1}{2}\)
Cho A= 1 + \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\)
Tìm dư khi chia số A cho 2013.
[4+6+8+10+...+2012]*1/1000*[1/2+3/4+5/6]
So sánh S và P biết:
S=\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
P=\(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
