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BL
0
BL
1
BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
XO
18 tháng 10 2020
Ta có \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}=\frac{a+a^2+....+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\frac{a}{a^2}=\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\)
=> \(\left(\frac{a}{a^2}\right)^{2020}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)
=> \(\frac{a}{a^2}.\frac{a}{a^2}...\frac{a}{a^2}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(2020 thừa số \(\frac{a}{a^2}\))
=> \(\frac{a}{a^2}.\frac{a^2}{a^3}...\frac{a^{2020}}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(Vì \(\frac{a}{a^2}=\frac{a^2}{a^3}=...=\frac{a^{2020}}{a^{2021}}\))
=> \(\frac{a}{a^{2021}}=\left(\frac{a+a^2+...+a^{2020}}{a^2+a^3+...+a^{2021}}\right)^{2020}\)(đpcm)
12 tháng 1 2022
{ x + y + z = 1 (1)
{ x² + y² + z² = 1 (2)
{ x³ + y³ + z³ = 1 (3)
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
⇒ 2(xy + yz + zx) = (x + y + z)² - (x² + y² + z²) = 1² - 1 = 0 ⇒ xy + yz + zx = 0
(x + y + z)³ = x³ + y³ + z³ + 3(x + y)(y + z)(z + x)
⇒ 3(x + y)(y + z)(z + x) = (x + y + z)³ - (x³ + y³ + z³) = 1³ - 1 = 0
⇒ x + y = 0 hoặc y + z = 0 hoặc z + x = 0
@ Nếu x + y = 0 ⇔ x = - y thay vào (1) ⇒ z = 1 , thay vào (2) ⇒ 2x² + 1 = 1 ⇒ x = 0; y = 0
⇒ S = 1
Tương tự cho trường hợp y + z = 0 và z + x = 0
DH
2
7 tháng 11 2023
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2021}}\)
=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}\)
=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2020}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2021}}\)
=>\(2A=1-\dfrac{1}{3^{2021}}\)
=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2021}}< \dfrac{1}{2}\)
12 tháng 12 2021
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) + ... + \(\dfrac{1}{3^{2020}}\) + \(\dfrac{1}{3^{2021}}\) < \(\dfrac{1}{2}\)
3.B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ ... + \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\)
3B - B = 1+\(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\) - (\(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ ... + \(\dfrac{1}{3^{2020}}\)+\(\dfrac{1}{3^{2021}}\))
2B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{3^{2019}}\) + \(\dfrac{1}{3^{2020}}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}\)- ...- \(\dfrac{1}{3^{2020}}\)-\(\dfrac{1}{3^{2021}}\)
2B = (1 - \(\dfrac{1}{3^{2021}}\)) + (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3^2}\) - \(\dfrac{1}{3^2}\)) +...+ (\(\dfrac{1}{3^{2020}}\) - \(\dfrac{1}{3^{2020}}\))
2B = 1 - \(\dfrac{1}{3^{2021}}\)
B = (1 - \(\dfrac{1}{3^{2021}}\)) : 2
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{2021}}\) < \(\dfrac{1}{2}\) (đpcm)