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1) Quy đồng
2) Rút gọn
3) Phân số trung gian
4) So sánh với 1
5) Dùng phần thiếu , phần thừa
1 ) Quy đồng
2 ) Rút gọn
3 ) Phân số trung gian
4 ) So sánh với 1
5 ) Dùng phần thiếu , phần thừa
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Số phàn tử:
\(2029-2021+1=9\)
Tổng dãy trên:
\(\left(2029+2021\right)\cdot\dfrac{9}{2}=18225\)
Số hạng là:
(2029-2021):1+1=9
Tổng là:(2029+2021).9:2=18225
Đáp số :18225
Chúc bạn học tốt nha
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A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2022}{50^8}\)
A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
B = \(\dfrac{2023}{50^{10}}\) + \(\dfrac{2021}{5^8}\) = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{1}{50^{10}}\) + \(\dfrac{2021}{50^8}\)
Vì: \(\dfrac{1}{50^{10}}\) < \(\dfrac{1}{50^8}\) nên \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^{10}}\) < \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
Vậy A > B
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\(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}\)
Ta thấy các phân số \(\frac{1}{101};\frac{1}{102};\frac{1}{103};...;\frac{1}{198};\frac{1}{199}\)đều lớn hơn \(\frac{1}{200}\)
\(\Rightarrow A>\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+..+\frac{1}{200}+\frac{1}{200}\)(có 100 số hạng \(\frac{1}{200}\))
\(\Leftrightarrow A>\frac{100}{200}\)
\(\Leftrightarrow A>\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)
\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)
\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)
\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:\(\frac{4}{7^{2020}}+\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{5^{60}+1}{5^{61}+1}\)
\(5A=\frac{5(5^{60}+1)}{5^{61}+1}=\frac{5^{61}+5}{5^{61}+1}=\frac{5^{61}+1+4}{5^{61}+1}=1+\frac{4}{5^{61}+1}\) \((1)\)
\(B=\frac{5^{61}+1}{5^{62}+1}\)
\(5B=\frac{5(5^{61})+1}{5^{62}+1}=\frac{5^{62}+5}{5^{62}+1}=\frac{5^{62}+1+4}{5^{62}+1}=1+\frac{4}{5^{62}+1}\) \((2)\)
Từ 1 và 2 \(\Rightarrow1+\frac{4}{5^{61}+1}>1+\frac{4}{5^{62}+1}\)
\(\Rightarrow5A>5B\)
Hay \(A>B\)
Vậy : ...
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a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)
\(x=\dfrac{4}{-10}+\dfrac{3}{5}\)
\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)
\(x=\dfrac{1}{5}\)
b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}-2+2=\dfrac{4}{x}+4+2\)
\(\dfrac{3}{x}=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)
\(3x=\left(4x+4\right)x\)
\(3x=5x\cdot x+4x\)
\(3x=x\left(5x+4\right)\)
\(3=5x+4\)
\(5x=-1\)
\(x=\dfrac{-1}{5}\)
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A = 2019 \(\times\) 2021 + 2023
A = (2018 + 1).(2022 -1) + 2023
A = 2018.2022 - 2018 + 2023 > 2018.2022 - 2022
Vậy A > B
Cách 1: Nhìn qua là biết A > B :))
Cách 2: Giải cụ thể:
A = 2019 x 2021 + 2023
= 2018 x 2021 + 2021 + 2023 = 2018 x 2021 + 4044
B = 2018 x 2022 - 2022
= 2018 x 2021 + 2018 - 2022 = 2018 x 2021 - 4
⇒ A > B và lớn hơn: 4044 + 4 = 4048
\(A=2019\times2021=\left(2021-1\right)\times\left(2021+1\right)=2021^2-1< 2021^2=B.\)
sai mất rồi nhưng dù sao cũng cảm ơn bn nhé