Tính giá trị biểu thức A biết:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
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Những câu hỏi liên quan
31 tháng 5 2018
\(A=16-\frac{\left(-2\right)\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)
\(A=16-\frac{-2}{\frac{1}{3}}=16-\left(-6\right)=22\)
Vậy A = 22
KN
12 tháng 8 2016
=1/10+1/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10
=1/10+45/10
=46/10=23/5
9 tháng 10 2019
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 3 2019
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
30 tháng 3 2022
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
24 tháng 3 2022
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
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HT
Ta có:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(\Rightarrow A=16+\frac{\frac{2}{9}+\frac{2}{10}+\frac{2}{11}+...+\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(\Rightarrow A=16+\frac{2\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)
\(\Rightarrow A=16+\frac{2}{\frac{1}{3}}\)
\(\Rightarrow A=16+\left(2:\frac{1}{3}\right)\)
\(\Rightarrow A=16+\left(2.3\right)\)
\(\Rightarrow A=16+6\)
\(\Rightarrow A=22\)
Vậy\(A=22\)
A = 16 + (2/9+2/10+....+2/2020)/(1/27+1/30+.....+1/6060)
= 16 + 6
= 22
Tk mk nha