3\(\dfrac{1}{2}\) - \(\dfrac{4}{5}\) - 8\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) - (2021)^0
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*Thực hiện1/ (dfrac{2021}{2020}+dfrac{2020}{2021}) x (dfrac{1}{2}-dfrac{1}{3}-dfrac{1}{6})2/ (dfrac{7}{19}-dfrac{5}{12}):dfrac{-5}{8}-(dfrac{7}{19}-dfrac{29}{12}):dfrac{5}{8}3/ dfrac{-5}{6}xdfrac{7}{24}-dfrac{5}{6}xdfrac{14}{24}-dfrac{5}{6}xdfrac{3}{24}
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*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
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mink chịu bài này nó rất khó
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3 \(\dfrac{1}{2}\) - \(\dfrac{4}{5}\) - \(8\dfrac{1}{2}\) - \(\dfrac{1}{5}\) - (2021)^0
\(=\left(3-8+\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-1\)
\(=-5-1-1=-7\)
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1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
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12 tháng 3 2023 lúc 17:48
Tính giá trị biểu thức: M= \(\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
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Câu 5 : A dfrac{1}{2} +dfrac{1}{2^2}+ dfrac{1}{2^3}+ dfrac{1}{2^4}+ ....+dfrac{1}{2^{2021}}+dfrac{1}{2^{2022}}và B dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{17}{60}a) Rút gọn Ab) So sánh A và B
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Câu 5 : A= \(\dfrac{1}{2}\) +\(\dfrac{1}{2^2}\)+ \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)+ ....+\(\dfrac{1}{2^{2021}}\)+\(\dfrac{1}{2^{2022}}\)và B= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{17}{60}\)
a) Rút gọn A
b) So sánh A và B
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
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a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )
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4 tháng 5 2022 lúc 20:30
Câu 1: Thực hiện phép tính
a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)
b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)
Câu 2: Tìm x
\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
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dfrac{2}{5} x dfrac{3}{4}-dfrac{1}{8}
dfrac{4}{3}+dfrac{1}{3}-dfrac{1}{5}dfrac{9}{20}-dfrac{3}{5} x dfrac{1}{4}
dfrac{2}{8}+dfrac{2}{3}:dfrac{4}{5}
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\(\dfrac{2}{5}\) x \(\dfrac{3}{4}-\dfrac{1}{8}\) \(= \)
\(\dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\)
\(\dfrac{9}{20}-\dfrac{3}{5}\) x \(\dfrac{1}{4}\)\(= \)
\(\dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}\)\(=\)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
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Chứng minh rằng :
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
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Giải:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
Ta có:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
Mà \(\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(\Rightarrow2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\left(đpcm\right)\)
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A 1dfrac{1}{2} times1dfrac{1}{3} times1dfrac{1}{4} times1dfrac{1}{5} times.... times1dfrac{1}{2020} times 1dfrac{1}{2021}
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A= 1\(\dfrac{1}{2}\) \(\times\)1\(\dfrac{1}{3}\) \(\times\)1\(\dfrac{1}{4}\) \(\times\)1\(\dfrac{1}{5}\) \(\times\).... \(\times\)1\(\dfrac{1}{2020}\) \(\times\) 1\(\dfrac{1}{2021}\)
\(A=1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times....\times1\dfrac{1}{2020}\times1\dfrac{1}{2021}\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times....\times\dfrac{2022}{2021}\\ =\dfrac{3\times4\times5\times6\times.....\times2022}{2\times3\times4\times5\times....\times2021}\\ =\dfrac{2022}{2}=1011\)
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