cho A =(\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2024^2}\))
So sánh A với \(\dfrac{1}{2}\)
K
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7 tháng 4 2023
Ta có :22A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)
22A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)
3A=\(\dfrac{\left(2n\right)^2-1}{\left(2n\right)^2}\) <\(\dfrac{n^2}{\left(2n\right)^2}\)=\(\dfrac{1}{2}\)
3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 4 2023
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)+ \(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+ \(\dfrac{1}{2^2.n^2}\)
A = \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{n^2}\))
22 \(\times\) A = 1 + \(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+......+\(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +......+ \(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)
1 = 1
\(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)
...................
\(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)
Cộng vế với vế ta có:
4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)
4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)
A < ( 2 - \(\dfrac{1}{n}\)): 4
A < 2 : 4 - \(\dfrac{1}{n}\) : 4
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)
Vậy A < \(\dfrac{1}{2}\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 9 2023
1) Ta thấy:
\(4=1+3=1+\sqrt{9}\)
\(1+2\sqrt{2}=1+\sqrt{2^2\cdot2}=1+\sqrt{8}\)
Mà: \(\sqrt{8}< \sqrt{9}\)
\(\Rightarrow1+\sqrt{8}< 1+\sqrt{9}\)
\(\Rightarrow\dfrac{1}{1+\sqrt{8}}>\dfrac{1}{1+\sqrt{9}}\)
\(\Rightarrow\dfrac{1}{1+2\sqrt{2}}>\dfrac{1}{4}\)
2) Ta thấy:
\(2018< 2024\)
\(\Rightarrow\sqrt{2018}< \sqrt{2024}\) (1)
\(2025< 2026\)
\(\Rightarrow\sqrt{2025}< \sqrt{2026}\) (2)
Từ (1) và (2) ta có:
\(\sqrt{2018}+\sqrt{2025}< \sqrt{2024}+\sqrt{2026}\)
31 tháng 10 2023
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
H
10 tháng 4 2021
a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)
Áp dụng bổ đề trên ta có:
\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)
2 tháng 5 2021
A=(1/2^2-1) (1/3^2-1) (1/4^2-1) .... (1/100^2-1)
A=(1/2^2-2^2/2^2) (1/3^2-3^2/3^2) ...... (1/100^2-100^2/100^2)
A=1-2^2/2^2 . 1-3^2/3^2 .... 1-100^2/100^2
A=-(2^2-1/2^2 . 3^2-1/3^2 ..... 100^2-1/100^2)
A=-(1.3/2^2 x 2.4/3^2 ..... 99.101/100^2)
A=-(1.3.2.4.....99.101/2.2.3.3.....100.100)
A=-[(1.2.3....99).(3.4.5.....101) / (2.3.4...100) . (2.3.4...100) ]
A=-101/200 < -1/2
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
MH
6 tháng 5 2022
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)