Tính tổng
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\)
\(B=\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{2015}{2016}\)
Những câu hỏi liên quan
Bài 1 : Tính :a)frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right)times230frac{1}{5}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right)divleft(12frac{1}{3}-14frac{2}{7}right)}b) frac{2^{12}times3^5-4^6times9^2}{left(2^4times3right)^6+8^4times3^5}-frac{5^{10}times7^3-25^5times49^2}{left(125times7right)^3+5^9times14^3}c)Pfrac{frac{1}{2}+frac{1}{3}+frac{1}{4}+....+frac{1}{2016}}{frac{2015}{1}+frac{2014}{2}+frac{2013}{3}+....+frac{1}{2015}}
Đọc tiếp
Bài 1 : Tính :
a)\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) \(\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^4\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3-25^5\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)
c)P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+....+\frac{1}{2015}}\)
Xem thêm câu trả lời
A =\(\frac{2015+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+\frac{2011}{5}+.....+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2016}}=\)
tìm A
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
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Cho :
A = \(\left(\frac{1}{2}_{ }+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2016}+\frac{1}{2017}\right)\)
B = \(\left(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Tính \(\frac{B}{A}\) ?
[Các bạn giúp mình với !!!]
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
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\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
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Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
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Tính:
a/\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
b/\(\frac{2\cdot2306}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+230}}\)c/\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44\cdot49}\right)\left(\frac{1-3-5-...-49}{89}\right)\)
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
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a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
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Tính nhanh : \(\frac{2017+\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}\)
tính
A=\(\left(\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}\right)\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
Cho biểu thức sau: \(P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+.....+\frac{2015}{5^{2015}}+\frac{2016}{5^{2016}}\)
Chứng minh 1/4 < P< 1/3
Thực hiện phép tính :
a)A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{2017}}{2017+\frac{2016}{2}+\frac{2015}{3}+.............\frac{1}{2016}}\)
b)B=\(\frac{-1^2}{1.2}.\frac{-2^2}{2.3}+..........+\frac{-100^2}{100.101}.\frac{-101^2}{101.102}\)
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