cho biết \(A=\frac{2016^2+1^2}{2016.1}+\frac{2015^2+2^2}{2015.2}+\frac{2014^2+3^2}{2014.3}+...+\frac{1009^2+1008^2}{1009.1008}\) ;B=\(\frac{1+1+1+1+...+1+1}{2+3+4+..+2017}\)tìm \(\frac{A}{B}\)
Những câu hỏi liên quan
Cho \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2016\right)}{2016.1+2015.2+2014.3+...+2.2015+1.2016}\)
Cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
P=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...........+\frac{1}{2014}+\frac{1}{2015}\)
Tính (S-P)2016
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)
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Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)
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1.So sánh:
\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}\) và \(4\)
2. Tính :
\(\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\)
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
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Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
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Tính \(\left(S-P\right)^{2015}+\left(S+P\right)^{2016}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2005}\)
\(P=\frac{1}{1008}+\frac{1}{1009}+.........+\frac{1}{2014}+\frac{1}{2015}\)
Giúp mình nhanh nhé!
a)so sánh:sqrt{17}+sqrt{26}+1vàsqrt{99}b)chứng minh:frac{1}{sqrt{ }1}+frac{1}{sqrt{ }2}+frac{1}{sqrt{ }3}+...+frac{1}{sqrt{ }99}+frac{1}{sqrt{ }100}10c)cho:S1-frac{1}{2}+frac{1}{3}-frac{1}{4}+...+frac{1}{2013}-frac{1}{2014}+frac{1}{2015}vàPfrac{1}{1008}+frac{1}{1009}+frac{1}{1010}+...+frac{1}{2014}+frac{1}{2015}tính left(S-Pright)^{2016}
Đọc tiếp
a)so sánh:
\(\sqrt{17}+\sqrt{26}+1và\sqrt{99}\)
b)chứng minh:\(\frac{1}{\sqrt{ }1}+\frac{1}{\sqrt{ }2}+\frac{1}{\sqrt{ }3}+...+\frac{1}{\sqrt{ }99}+\frac{1}{\sqrt{ }100}>10\)
c)cho:S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)vàP=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)tính \(\left(S-P\right)^{2016}\)
\(A=\frac{[(\frac{2}{1001}-\frac{3}{2002})\frac{1001}{17}+\frac{33}{34}]}{[(\frac{7}{1008}+\frac{11}{2016})\frac{1008}{25}+\frac{1009}{2016}]}\)
A=\(\frac{[\left(\frac{2}{1000}-\frac{3}{2002}\right)\frac{1001}{17}+\frac{33}{44}]}{[\left(\frac{7}{1008}+\frac{11}{2016}\right)\frac{1008}{25}+\frac{1009}{2016}]}\)
Câu hỏi:
Cho M=
1+(1+2)+(1+2+3)+…+(1+2+3+…+2016)
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2016.1+2015.2+2014.3+…+2.2015+1.2016
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