Cho \(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........-\frac{1}{2014}\)va \(D=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+........+\frac{1}{2014}\)tinh \(\left(\frac{C}{D}\right)^{2015}\)
Những câu hỏi liên quan
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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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 \(\left(S-P\right)^{2016}\)
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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CMR :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)
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a ) S = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\) và P = \(\frac{1}{1008}\) + \(\frac{1}{1009}+\frac{1}{1010}+........+\frac{1}{2014}+\frac{1}{2015}\)
Tính (S-P)^2016.
b, Tìm x,y biết : |x - 5 | + |1- x | = \(\frac{12}{\left|y+1\right|+3}\)
c, Tìm số tự nhiên x thoả mãn : \(3^x+4^x=5^x\)
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!
Bài 1: Tính giá trị biểu thức A frac{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2018}}{frac{2017}{1}+frac{2016}{2}+frac{2015}{3}+...+frac{1}{2017}}B frac{frac{1}{51}+frac{1}{53}+frac{1}{55}+...+frac{1}{149}}{frac{1}{101.99}+frac{1}{103.97}+...+frac{1}{149.51}}C frac{1+frac{1}{3}+frac{1}{5}+...+frac{1}{99}}{frac{1}{1.99}+frac{1}{3.97}+...+frac{1}{99.1}}D frac{frac{1}{1.300}+frac{1}{2.301}+frac{1}{3.302}+...+frac{1}{101.400}}{frac{1}{1.102}+frac{1}{2.103}+frac{1}{3.104}+...+frac{1}{299.4000}} ...
Đọc tiếp
Bài 1: Tính giá trị biểu thức
A= \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
B= \(\frac{\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)
C= \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)
D= \(\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.4000}}\)
Bài 2: Tìm x, biết:
a) \(\frac{x+1}{2014}+\frac{x+2}{2013}+...+\frac{x+1007}{1008}=\frac{x+1008}{1007}+\frac{x+1009}{1006}+...+\frac{x+2014}{1}\)
b) \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-1}{4}\)
Các bạn làm hết giúp mik nha! ^ ^
Bài 1 :
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\left(\frac{2017}{1}+1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2018}{1}+\frac{2018}{2}+\frac{2018}{3}+....+\frac{2018}{2017}+\frac{2018}{2018}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)}\)
\(=\frac{1}{2018}\)
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B=\(\frac{\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)
\(\)TA CÓ E=\(\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}\)
\(200E=\frac{200}{101.99}+\frac{200}{103.97}+..+\frac{200}{149.51}\)
\(200E=\frac{101+99}{101.99}+\frac{103+97}{103.97}+...+\frac{149+51}{149.51}\)
\(200E=\frac{1}{99}+\frac{1}{101}+\frac{1}{97}+\frac{1}{103}+...+\frac{1}{51}+\frac{1}{149}\)
\(200E=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\)
\(E=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right):200\)\(=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}\)/\(\left(\frac{1}{51}+\frac{1}{53}+..+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{\frac{1}{200}}=200\)
VẬY B=200
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Còn câu C thì bạn làm tương tự câu B thôi bạn nhé
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Chứng tỏ:a) frac{2013}{2014}+frac{2014}{2015}+frac{2015}{2013}3b) left(1+frac{1}{2}right)left(1+frac{1}{2^2}right)left(1+frac{1}{2^3}right)left(1+frac{1}{2^4}right)....left(1+frac{1}{2^{50}}right) 3c) Cfrac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{9999}{10000} frac{1}{100}d) frac{1}{2}-frac{1}{2^2}+.............+frac{1}{2^{99}}-frac{1}{2^{100}} frac{1}{3}
Đọc tiếp
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)....\left(1+\frac{1}{2^{50}}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.............+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có :
\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)
\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)
\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)
\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)
Mà :
\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)
Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế )
\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 )
\(\Rightarrow\)\(A>3\) ( điều phải chứng minh )
Vậy \(A>3\)
Chúc đệ học tốt ~
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c,
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
vì \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.............................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
bt lm mỗi một câu :v
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Xem thêm câu trả lời
Chứng tỏ:a) frac{2013}{2014}+frac{2014}{2015}+frac{2015}{2013}3b) left(1+frac{1}{2}right)left(1+frac{1}{2^2}right)left(1+frac{1}{2^3}right)left(1+frac{1}{2^4}right)...left(1+frac{2}{50}right) 3c) Cfrac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{9999}{10000} frac{1}{100}d) frac{1}{2}-frac{1}{2^2}+.........+frac{1}{2^{99}}-frac{1}{2^{100}} frac{1}{3}
Đọc tiếp
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)...\left(1+\frac{2}{50}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.........+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
,mình sửa lại đề:
\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)
xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015
=\(\frac{2013}{2013}\)
=\(1\)
vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
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