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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
Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)
Được cập nhật 41 giây trước (17:23)
Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1
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\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
\(C=1-2+2^2-2^3+...-2^{2011}+2^{2012}\)
\(\Rightarrow2C=2-2^2+2^3-2^4+...-2^{2012}+2^{2013}\)
\(\Rightarrow3C=1+2^{2013}\)
\(\Rightarrow C=\frac{1+2^{2013}}{3}\)
Vậy
\(D=-2+2^2-2^3+2^4-...-2^{2019}+2^{2020}\)
\(\Rightarrow-2D=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(\Rightarrow-3D=-2^{2021}+2\)
\(\Leftrightarrow D=\frac{2^{2021}-2}{3}\)
a) 10 + (-2020) + (-16) + 2020 + 2
= 10 + [(-2020) + 2020] + (-16) + + 2
= 10 + 0 + (-16) + 2
= 10 + (-16) + 2
= (-6) + 2
= -4
b) (-123) + 54 + (-24) + 123 - 52 : 25
= (-123) + 54 + (-24) + 123 - 25 : 25
= (-123) + 54 + (-24) + 123 - 1
= [(-123) + 123] + 54 + (-24) - 1
= 0 + 54 + (-24) - 1
= 54 + (-24) - 1
= 30 - 1
= 29