\(\dfrac{2021.2023-1}{2019+2021.2022}\)
\(\dfrac{2022.2023-1}{2022.2023}và\dfrac{2021.2022-1}{2021.2022}\)
\(\dfrac{2022.2023}{2022.2023+1}và\dfrac{2023.2024}{2023.2024+1}\)
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:
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+,.......+\(\dfrac{1}{2021.2022}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
Tính
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{2021.2023}\)
Ta có :
\(\dfrac{1}{1.3}\text{=}2\left(\dfrac{1}{1}-\dfrac{1}{3}\right)\)
\(\dfrac{1}{3.5}\text{=}2\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(\dfrac{1}{5.7}\text{=}2\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(...\)
\(\dfrac{1}{2021.2023}\text{=}2\left(\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(\Rightarrow\) biểu thức chỉ còn :
\(2.1-\dfrac{2}{2023}\text{=}\dfrac{4044}{2023}\)
đặt biểu thức trên là A
ta có
2A=2/1.3+2/3.5+...+2/2021.2023
2A=1/1-1/3+1/3-1/5+...+1/2021-1/2023
2A=1/1-1/2023
2A=2022/2023
A=(2022/2023):2
A=1011/2023
Thực hiện phép tính ( tính hợp lý nếu có thể)
\(1-\dfrac{2}{1.3}-\dfrac{2}{3.5}-..\dots-\dfrac{2}{2021.2023}\)
Tính nhanh:
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)
\(A=1.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{5}+...+\dfrac{1}{2021}.\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2021\cdot2023}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
\(x-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{2021.2022}-\dfrac{1}{2022.2023}=\dfrac{-2024}{2023}\)
x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023
x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023
x-(1-1/2023)=-2024/2023
x-2022/2023=-2024/2023
x = -2024/2023+2022/2023
x = -2/2023
Vậy x = -2/2023
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
bài 4 : tính
A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ...+ \(\dfrac{1}{2021.2023}\)
mọi người giải giúp em bài này nha
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
\(\dfrac{7}{1.3}\) + \(\dfrac{7}{3.5}\) + \(\dfrac{7}{5.7}\) + ... + \(\dfrac{7}{2021.2023}\)
Đặt tông trên là A
\(\dfrac{2A}{7}=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(\Rightarrow A=\dfrac{7.2022}{2.2023}=\dfrac{1011}{289}\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{2021.2023}\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2023-2021}{2021.2023}\)
\(=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{2023}{2021.2023}-\dfrac{2021}{2021.2023}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}...+\dfrac{2}{2021.2023}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}\)
\(=\dfrac{2023}{2023}-\dfrac{1}{2023}\)
\(=\dfrac{2022}{2023}\)
Tính nhanh: A= \(\dfrac{2022}{1.2}+\dfrac{2022}{2.3}+\dfrac{2022}{3.4}+...+\dfrac{2022}{2021.2022}\)
A=2022(1/1-1/2+1/2-1/3+...+1/2021-1/2022)
=2022(1/1-1/2022)
=2022.2021/2022
ket qua tu tinh nha
A = \(\dfrac{2022}{1.2}+\dfrac{2022}{2.3}+\dfrac{2022}{3.4}+...+\dfrac{2022}{2021.2022}\)
= \(\dfrac{2022}{1}-\dfrac{2022}{2}+\dfrac{2022}{2}-\dfrac{2022}{3}+\dfrac{2022}{3}-\dfrac{2022}{4}+...+\dfrac{2022}{2021}-\dfrac{2022}{2022}\)
= \(\dfrac{2022}{1}-\dfrac{2022}{2022}\)
= \(2021\)
Chúc bạn học tốt!! ^^
2022.( 1/1.2+ 1/2.3+...+ 1/2021+2022)
2022. ( 1-1/2+1/2-1/3+......+1/2021-1/2022)
2022.( 1-1/2022)
2022.2021/2022
2021