Ta có: \(1+\frac12\left(1+2\right)+\frac13\left(1+2+3\right)+\cdots+\frac{1}{2023}\left(1+2+\cdots+2023\right)\)

\(=1+\frac12\cdot\frac{2\cdot3}{2}+\frac13\cdot\frac{3\cdot4}{2}+...+\frac{1}{2023}\cdot\frac{2023\cdot2024}{2}\)

\(=1+\frac32+\frac42+\cdots+\frac{2024}{2}=\frac12\left(2+3+4+\cdots+2024\right)\)

\(=\frac12\left(2024-2+1\right)\cdot\frac{\left(2024+2\right)}{2}=\frac12\cdot2023\cdot\frac{2026}{2}=\frac{2023}{2}\cdot1013\)

=\(\frac{2049299}{2}\)

Ta có: \(2023-\frac12\left(1+2\right)-\frac13\left(1+2+3\right)-\cdots-\frac{1}{2022}\left(1+2+\cdots+2022\right)\)

\(=2023-\frac12\cdot\frac{2\cdot3}{2}-\frac13\cdot\frac{3\cdot4}{2}-\cdots-\frac{1}{2022}\cdot\frac{2022\cdot2023}{2}\)

\(=2023-\frac32-\frac42-\cdots-\frac{2023}{2}=2023-\frac12\left(3+4+\cdots+2023\right)\)

\(=2023-\frac12\frac{\left(2023-3+1\right)\left(2023+3\right)}{2}=2023-\frac12\cdot\frac{2021\cdot2026}{2}=2023-\frac12\cdot2021\cdot1013\)

=-1021613,5

a: =-3/4-1/4+2/7+5/7+2023/2024

=-1+1+2023/2024=2023/2024

b: 2/3x=2/7

=>x=2/7:2/3=3/7

c; =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5*3/2=9/10

Lời giải:

\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)

\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)

\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)

\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)

\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)

\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)

\(\dfrac{1012}{2023}\)

a: \(\dfrac{1}{7}\cdot\dfrac{3}{8}+\dfrac{1}{7}\cdot\dfrac{5}{8}+\dfrac{\left(-1\right)^{2023}}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{3}{8}+\dfrac{5}{8}\right)-\dfrac{1}{7}\)

\(=\dfrac{1}{7}-\dfrac{1}{7}=0\)

b: \(-3-\dfrac{16}{23}-\sqrt{\dfrac{4}{49}}-\dfrac{7}{23}+\dfrac{\left(-3\right)^2}{7}\)

\(=-3-\left(\dfrac{16}{23}+\dfrac{7}{23}\right)-\dfrac{2}{7}+\dfrac{9}{7}\)

\(=-3-\dfrac{23}{23}+\dfrac{7}{7}\)

=-3-1+1

=-3

c: \(\dfrac{4^2\cdot0,2^3}{2^6}\)

\(=\dfrac{2^4\cdot0,008}{2^6}=\dfrac{0.008}{4}=0.002\)

Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)

\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)

=C

=>B-C=0

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

\(7832\)

\(B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...-\dfrac{1}{2022}+\dfrac{1}{2023}\\ \Rightarrow B=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)

\(\Rightarrow B=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\\ \Rightarrow2^2B=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)

\(\Rightarrow4B-B=\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\\ \Rightarrow3B=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)

\(\Rightarrow3B=1-\dfrac{3}{2^{2024}}\\ \Rightarrow B=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)

\(\Rightarrow B=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\\ B=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)

4072299/4048

cho mik câu trả lời cụ thể đc k bn