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\(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\)
\(\Rightarrow7A=7^{2023}-7^{2022}+7^{2021}-...+7^3-7^2\)
\(\Rightarrow8A=A+7A=7^{2022}-7^{2021}+...+7^2-7+7^{2023}-7^{2022}+...+7^3-7^2=7^{2023}-7\)
\(\Rightarrow A=\dfrac{7^{2023}-7}{8}\)
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
số năm nay luôn
Ta có: 1+2-3-4+5+6-7-8+.....-2019-2020+2021+2022
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
Sửa đề: 1-2-3+4+5-6-7+8+...-2018-2019+2020+2021-2022-2023
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+(2021-2022-2023)
=0+0+...+0+(-1-2023)
=-2024
\(A=7^{2024}-7^{2023}+7^{2022}-7^{2021}+...+7^2-7\)
=>\(7A=7^{2025}-7^{2024}+7^{2023}-7^{2022}+...+7^3-7^2\)
=>\(7A+A=7^{2025}-7^{2024}+7^{2023}-7^{2022}+...+7^3-7^2+7^{2024}-7^{2023}+...+7^2-7\)
=>\(8A=7^{2025}-7\)
=>\(A=\dfrac{7^{2025}-7}{8}\)
Ta có:
\(A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}\) và \(B=\frac{1}{7^{2019}}\)
Ta xét 2 trường hợp:
\(TH1:\frac{4-7^{2020}}{7^{2020}}=\frac{-7^{2020}+4}{7^{2020}}=-1+\frac{4}{7^{2020}}\)
\(TH2:\frac{5+7^{2021}}{7^{2021}}=1+\frac{5}{7^{2021}}\)
\(\Rightarrow\left(-1+\frac{4}{7^{2020}}\right)+\left(1+\frac{5}{7^{2021}}\right)\)
\(\Rightarrow\frac{4}{7^{2020}}+\frac{5}{7^{2021}}\)
\(Do:\)
\(\frac{4}{7^{2020}}>\frac{1}{7^{2019}}\)
\(\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)
Nên:\(\frac{4}{7^{2020}}+\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)
\(\Rightarrow A>B\)
Sửa đề: \(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\)
=>\(7A=7^{2023}-7^{2022}+7^{2021}-7^{2020}+...+7^3-7^2\)
=>\(8A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7+7^{2023}-7^{2022}+...+7^3-7^2\)
=>\(8A=7^{2023}-7\)
=>\(A=\dfrac{7^{2023}-7}{8}\)