Cho \(A=1+\dfrac{2^2}{3^2}+\dfrac{2^2}{5^2}+\dfrac{2^2}{7^2}+....+\dfrac{2^2}{2009^2}\)
So sánh A và 3
cho A =\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) so sánh A với 1
Bạn tham khảo lời giải tại đây:
https://olm.vn/hoi-dap/detail/81621153379.html
So sánh:
a) 430 và 3.2410
b) \(\dfrac{3}{1^2.2^2}\) + \(\dfrac{5}{2^2.3^2}\) + \(\dfrac{7}{3^2.4^2}\) +...+\(\dfrac{19}{9^2.10^2}\) và 1
a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)
\(=3^{11}\cdot2^{30}\)
\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)
Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)
Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)
b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Vậy dãy trên nhỏ hơn 1
a/
\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)
\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)
\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)
b/
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)
\(=1-\dfrac{1}{10^2}< 1\)
a) 4³⁰ = (2²)³⁰ = 2⁶⁰ = 2³⁰.2³⁰ = 1073741824.2³⁰
3.24¹⁰ = 3.(3.2³)¹⁰ = 3.3¹⁰.2³⁰ = 3¹¹.2³⁰ = 177147.2³⁰
Do 1073741824 > 177147
⇒ 1073741824.2³⁰ > 177147.2³⁰
Vậy 4³⁰ > 3.24¹⁰
b) 3/(1².2²) + 5/(2².3²) + ... + 19/(9².10²)
= 1/1² - 1/2² + 1/2² - 1/3² + ... + 1/9² - 1/10²
= 1 - 1/100
= 99/100
Mà 99/100 < 1
⇒ 3/(1².2²) + 5/(2².3²) + 7/(3².4²) + ... + 19/(9².10²) < 1
Câu 5 : A= \(\dfrac{1}{2}\) +\(\dfrac{1}{2^2}\)+ \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)+ ....+\(\dfrac{1}{2^{2021}}\)+\(\dfrac{1}{2^{2022}}\)và B= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{17}{60}\)
a) Rút gọn A
b) So sánh A và B
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) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )
Cho A = 40 + \(\dfrac{3}{8}+\dfrac{7}{8^2}+\dfrac{5}{8^3}+\dfrac{32}{8^5}\)
B = \(\dfrac{24}{8^2}+40+\dfrac{5}{8^2}+\dfrac{40}{8^4}+\dfrac{5}{8^4}\)
So sánh A và B
a) So sánh các phân số:
\(\dfrac{2}{5}\) và \(\dfrac{2}{7}\); \(\dfrac{5}{9}\) và \(\dfrac{5}{6}\); \(\dfrac{11}{2}\) và \(\dfrac{11}{3}\).
b) Nêu cách so sánh hai phân số có cùng tử số.
a
2/5> 2/7
5/9<5/6
11/2>11/3
cách so sánh :
sét mẫu số của phân số này bé hơn mẫu số của phân số kia thì phân số này lớn hơn
mẫu số của phân số này lớn hơn mẫu số của phân số kia thì phân số này bé hơn
So sánh các phân số sau
\(a,\dfrac{-7}{6}và\dfrac{-11}{9}\) b,\(\dfrac{5}{-7}và\dfrac{-4}{5}\)
c,\(\dfrac{-8}{7}và\dfrac{-2}{5}\) d,\(\dfrac{-2}{5}và\dfrac{1}{3}\)
a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)
\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)
mà -21>-22
nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)
b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)
\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
mà -25>-28
nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)
c: \(\dfrac{-8}{7}< -1\)
\(-1< -\dfrac{2}{5}\)
Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)
d: \(-\dfrac{2}{5}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)
So Sánh : A = \(\dfrac{2009^{2009}+1}{2009^{2010}+1}\) và B = \(\dfrac{2009^{2010}-2}{2009^{2011}-2}\)
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
a) rút gọn: \(\dfrac{4^5x9^4-2x6^9}{2^{10}x3^8+6^8x20}\)
b) Cho A=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\).So sánh A với 2
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
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.