K=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+........+\left(1+2+3+.....+2012\right)}{2012.1+2011.2+2010.3+....+2.2011+1.2012}\)
Những câu hỏi liên quan
Cho K = 1+[1+2]+[1+2+3]+.....+[1+2+3+....+2012]/2012.1+2011.2+2010.3+....+2.2011+1.2012
Tính K+2011
Tính K=
1+(1+2)+(1+2+3)+(1+2+3+4)+....+(1+2+..+2012)/2012.1+2011.2+...1.2012
= 12 hoặc 13
các bạn cho mình vài li-ke cho tròn 480 với
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1) Tìm nghiệm nguyên của phương trình : xy+y = x3 +x2 +7
2) Giải phương trình : \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{2.2011}{2012}\)
1/
\(y\left(x+1\right)-x^2\left(x+1\right)=7\Leftrightarrow\left(x+1\right)\left(y-x^2\right)=7\)
TH1: \(\left\{{}\begin{matrix}x+1=1\\y-x^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=7\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+1=7\\y-x^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=37\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x+1=-1\\y-x^2=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x+1=-7\\y-x^2=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=63\end{matrix}\right.\)
2/
\(\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)...\left(1+\dfrac{1}{\left(x+1-1\right)\left(x+1+1\right)}\right)=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{\left(x+1\right)^2}{x\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{2.3.4...\left(x+1\right)}{3.4.5...\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\left(x+2\right)}=\dfrac{2.2011}{2012}\)
\(\Leftrightarrow2012\left(x+1\right)=2011\left(x+2\right)\)
\(\Leftrightarrow x=2010\)
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Cho K:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+........+\left(1+2+3+........+2012\right)}{2012\text{x}1+2011\text{x}2+2010\text{x}3+.............+1\text{x}2012}\)
Tính K cộng với 2011
cho K=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2012\right)}{2012\times1+2011\times2+2010\times3+...+2\times2011+1\times2012}\)
Tính k +2011
giúp nha -.- mk cho 5 tk
* Xét tử số của K, ta nhận thấy:
Số 1 được lấy 2012 lần
Số 2 được lấy 2011 lần
Số 3 được lấy 2010 lần
........
Số 2011 được lấy 2 lần
Số 2012 được lấy 1 lần
Vậy tử số viết được thành: 2012x1+2011x2+2010x3+...+2x2011+1x2012
Nên \(K=1\)
\(=>\)\(K+2011=2012\)
Vậy \(K+2011=2012\)
Chắc chắn đúng nhé!!
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k=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2013\right)}{2013\cdot1+2012\cdot2+2011\cdot3+...+2\cdot2012+1\cdot2013}\)
Tính K + 2013, biết:
K=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2013\right)}{2013\times1+2012\times2+2011\times3+...+2\times2012+1\times2013}\)
Tính P=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2012}\left(1+2+...+2012\right)\)
\(P=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(3+1\right).3}{2}+...+\frac{1}{2012}.\frac{\left(2012+1\right).2012}{2}\)
\(=1+\frac{\left(1+2\right)}{2}+\frac{\left(1+3\right)}{2}+...+\frac{\left(1+2012\right)}{2}\)
\(=1+\frac{2011}{2}+\frac{\left(2012+2\right).2011}{2}=1+\frac{2011}{2}+2011.1007\)
Rút gọn :
a/ \(A=\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}\)
b/ \(B=\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right)...\left(1+\frac{2012}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{2012}\right)}\)
