\(x-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{2021.2022}-\dfrac{1}{2022.2023}=\dfrac{-2024}{2023}\)
\(\dfrac{2022.2023-1}{2022.2023}và\dfrac{2021.2022-1}{2021.2022}\)
\(\dfrac{2022.2023}{2022.2023+1}và\dfrac{2023.2024}{2023.2024+1}\)
SO SÁNH
a, A = \(\dfrac{2022.2023-1}{2022.2023}\) = \(\dfrac{2022.2023}{2022.2023}\) - \(\dfrac{1}{2022.2023}\) = 1 - \(\dfrac{1}{2022.2023}\)
B = \(\dfrac{2021.2022-1}{2021.2022}\) = \(\dfrac{2021.2022}{2021.2022}\) - \(\dfrac{1}{2021.2022}\) = 1 - \(\dfrac{1}{2021.2022}\)
Vì \(\dfrac{1}{2022.2023}\) < \(\dfrac{1}{2021.2022}\)
Nên A > B
b, C = \(\dfrac{2022.2023}{2022.2023+1}\)
C = \(\dfrac{2022.2023+1-1}{2022.2023+1}\) = \(\dfrac{2022.2023+1}{2022.2023+1}\) - \(\dfrac{1}{2022.2023+1}\)
C = 1 - \(\dfrac{1}{2022.2023+1}\)
D = \(\dfrac{2023.2024}{2023.2024+1}\) = \(\dfrac{2023.2024+1-1}{2023.2024+1}\)
D = 1 - \(\dfrac{1}{2023.2024+1}\)
Vì \(\dfrac{1}{2022.2023+1}\) > \(\dfrac{1}{2023.2024+1}\)
Nên C < D
Tính tổng :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{2020.2021}+\dfrac{1}{2021.2022}\)
Dấu chấm là dấu nhân
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
Tính tổng:
\(M=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\)
\(M=\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(M=1-\dfrac{1}{7}\)
\(M=\dfrac{6}{7}\)
tham khảo
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
vào đi
refer
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
\(\text{Tìm x, biết:}\)
\(a\)) \(x-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-\dfrac{2}{9.11}-\dfrac{2}{11.13}-\dfrac{2}{13.15}=\dfrac{2}{5}\)
\(b\)) \(\dfrac{1}{2.3}.x+\dfrac{1}{3.4}.x+\dfrac{1}{4.5}.x+...+\dfrac{1}{49.50}.x=1\)
\(c\)) \(x-\dfrac{20}{11.3}-\dfrac{20}{13.15}-...-\dfrac{53}{55}=\dfrac{3}{11}\)
\(d\)) \(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\)
\(e\)) \(\left(\dfrac{11}{12}.\dfrac{11}{2.23}.\dfrac{11}{23.34}...\dfrac{11}{89.100}\right).x=\dfrac{5}{3}\)
\(f\)) \(\left(\dfrac{2}{11.13}.\dfrac{2}{13.15}.\dfrac{2}{15.17}...\dfrac{2}{19.21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}\) .-.
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{19.20}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=1-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
a) \(\dfrac{1}{2.3}x+\dfrac{1}{3.4}x+\dfrac{1}{4.5}x+....+\dfrac{1}{49.50}x=1\)
b) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{55}\)
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
tính nhanh
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}=?????\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
=\(1-\dfrac{1}{5}\)
=\(\dfrac{4}{5}\)
1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5
= 1/1 - 1/5
= 4/5
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)