chứng minh rằng :
b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
mọi người ơi chú ý hộ mik là cái chỗ \(\dfrac{1}{2014}\) trên kia là đúng nha
Những câu hỏi liên quan
25 tháng 7 2021 lúc 14:43
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
25 tháng 7 2021 lúc 15:02
Chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014^2}>\dfrac{1}{5}\)
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{150}=\dfrac{1}{5}\) (đpcm)
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25 tháng 7 2021 lúc 11:54
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
Cho A = \(\dfrac{2015}{2014^2+1}+\dfrac{2015}{2014^2+2}+\dfrac{2015}{2014^3+3}+....+\dfrac{2015}{2014^2+2014}\)
Chứng minh rằng A không là số nguyên dương
Các bạn ơi , giúp mình với T T
Thực hiện phép tính
a) A= \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)\)\(+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)
b) B=\(\dfrac{1-3}{1.3}+\dfrac{2-4}{2.4}+\dfrac{3-5}{3.5}+\dfrac{4-6}{4.6}+...+\dfrac{2011-2013}{2011.2013}+\dfrac{2012-2014}{2012.2014}-\dfrac{2013+2014}{2013.2014}\)
Cho A dfrac{1}{2014}+dfrac{2}{2013}+dfrac{3}{2012}+...+dfrac{2013}{2}+2014 B dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2015} Tính giá trị dfrac{A}{B}
Đọc tiếp
Cho A = \(\dfrac{1}{2014}\)+\(\dfrac{2}{2013}\)+\(\dfrac{3}{2012}\)+...+\(\dfrac{2013}{2}\)+2014
B = \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2015}\)
Tính giá trị \(\dfrac{A}{B}\)
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
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Tính
\(A=\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}+1\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}\right)-\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}+1\right)\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
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\(A=\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}+1\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}\right)-\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}+1\right)\)
tất nhên là bằng 00000000000000000000000000000000000000
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Tính giá trị của biểu thức:
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{1+\left(\dfrac{1}{2013}+1\right)+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{3}{2011}+1\right)+...+\left(\dfrac{2012}{2}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{2014}{2014}+\dfrac{204}{2013}+\dfrac{2014}{2012}+\dfrac{2014}{2011}+...+\dfrac{2014}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}=2014\)
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mình ko chắc đúng nha !
Số số hạng của tử là :
(2013-1):1+1=2013(số hạng)
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+.....+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{1}{2013}+1+\dfrac{2}{2012}+1+....+\dfrac{2012}{2}+1+\dfrac{2013}{1}-2012}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{2014}{2013}+\dfrac{2014}{2012}+....+\dfrac{2014}{2}+1}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=2014\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\right)\)
=2014
Mình ghi thêm ở cái dâu bằng thứ 2 cuối cùng trên tử có ghi trừ 2012 là do tử có 2013 hạng tử mà mình chỉ cộng 1 cho 2012 hạng tử nên phải trừ đi 2012
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