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Trả lời:
\(A=-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-\frac{1}{5^2}-...-\frac{1}{99^2}-\frac{1}{100^2}\)
\(=-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\right)\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
........
\(\frac{1}{99^2}< \frac{1}{98.99}\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\right)>-1\)
Vậy A > - 1
\(A=-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Ta có \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}< 1\)
=> A > -1
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1
gọi
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)
VÌ \(\frac{2019}{2020}< 1\Rightarrow A< 1\)
VẬY \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)
1. a) P = 4 - ( x - 2 )32
( x - 2 )32 ≥ 0 ∀ x => - ( x - 2 )32 ≤ 0 ∀ x
=> 4 - ( x - 2 )32 ≤ 4 ∀ x
Dấu bằng xảy ra <=> x - 2 = 0 => x = 2
Vậy PMax = 4 khi x = 2
b) Q = 20 - | 3 - x |
| 3 - x | ≥ 0 ∀ x => - | 3 - x | ≤ 0 ∀ x
=> 20 - | 3 - x | ≤ 20 ∀ x
Dấu bằng xảy ra <=> 3 - x = 0 => x = 3
Vậy QMax = 20 khi x = 3
c) C = \(\frac{5}{\left(x-3\right)^2+1}\)
Để C có GTLN => ( x - 3 )2 + 1 nhỏ nhất dương
=> ( x - 3 )2 + 1 = 1
=> ( x - 3 )2 = 0
=> x - 3 = 0
=> x = 3
=> CMax = \(\frac{5}{\left(3-3\right)^2+1}=\frac{5}{1}=5\)khi x = 3
Answer :
\(\Rightarrow A+1=1+1+2+2^2+...+2^{2021}\)
\(\Rightarrow A+1=2+2+2^2+...+2^{2021}\)
\(\Rightarrow A+1=2^2+2^2+2^3+...+2^{2021}\)
\(\Rightarrow A+1=2^3+2^3+2^4+...+2^{2021}\)
....
\(\Rightarrow A+1=2^{2021}+2^{2021}=2^{2022}\)
Mà \(2^x=A+1\Rightarrow2^x=2^{2022}\Rightarrow x=2022\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)
\(A=\left(3A-A\right):2\)
\(3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)
\(A=\left(3-\frac{1}{3^{2014}}\right):2\)
\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}\)
\(\Rightarrow A<\frac{3}{2}\)
Trả lời:
\(P=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2021^2}-1\right)\)
\(=\frac{1-2^2}{2^2}\cdot\frac{1-3^2}{3^2}\cdot\frac{1-4^2}{4^2}\cdot...\cdot\frac{1-2021^2}{2021^2}\)
\(=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-4084440}{2021^2}\)
\(=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{4084440}{2021^2}\) ( vì tích trên có 2020 thừa số, mà tích của 2020 thừa số âm là số dương )
\(=\frac{3\cdot8\cdot15\cdot...\cdot4084440}{2^2\cdot3^2\cdot4^2\cdot...\cdot2021^2}\)
\(=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2020\cdot2022}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2021\cdot2021}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot2020\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot2022\right)}{\left(2\cdot3\cdot4\cdot...\cdot2021\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot2021\right)}\)
\(=\frac{1\cdot2022}{2021\cdot2}=\frac{1011}{2021}>\frac{1011}{2022}=\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Kết luận B<1/2