Những câu hỏi liên quan
Mark
Xem chi tiết
Linh_Windy
1 tháng 10 2017 lúc 13:58

\(LINH=\dfrac{3}{1^2.2^2}+\dfrac{7}{3^2.4^2}+\dfrac{11}{5^2.6^2}+\dfrac{15}{7^2.8^2}+\dfrac{19}{9^2.10^2}\)

\(LINH=\dfrac{1+2}{1^2.2^2}+\dfrac{3+4}{3^2.4^2}+\dfrac{5+6}{5^2.6^2}+\dfrac{7+8}{7^2.8^2}+\dfrac{9+10}{9^2.10^2}\)

\(LINH=\dfrac{1}{1^2.2^2}+\dfrac{2}{1^2.2^2}+\dfrac{3}{3^2.4^2}+\dfrac{4}{3^2.4^2}+\dfrac{5}{5^2.6^2}+\dfrac{6}{5^2.6^2}+\dfrac{7}{7^2.8^2}+\dfrac{8}{7^2.8^2}+\dfrac{9}{9^2.10^2}+\dfrac{10}{9^2.10^2}\)

\(LINH=\dfrac{1}{1.2^2}+\dfrac{1}{1^2.2}+\dfrac{1}{3.4^2}+\dfrac{1}{3^2.4}+\dfrac{1}{5.6^2}+\dfrac{1}{5^2.6}+\dfrac{1}{7.8^2}+\dfrac{1}{7^2.8}+\dfrac{1}{9.10^2}+\dfrac{1}{9^2.10}\)\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+\dfrac{1}{180}+\dfrac{1}{150}+\dfrac{1}{448}+\dfrac{1}{392}+\dfrac{1}{900}+\dfrac{1}{810}\)Vì:

\(\left\{{}\begin{matrix}\dfrac{1}{48}< \dfrac{1}{32}\\\dfrac{1}{36}< \dfrac{1}{32}\\...............\\\dfrac{1}{810}< \dfrac{1}{32}\end{matrix}\right.\)

Nên:

\(\dfrac{1}{48}+\dfrac{1}{36}+.....+\dfrac{1}{810}< \dfrac{1}{32}+\dfrac{1}{32}+....+\dfrac{1}{32}\)

\(\Rightarrow\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{32}.8=\dfrac{1}{4}\)

Nên:

\(LINH=\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{48}+\dfrac{1}{36}+....+\dfrac{1}{810}< \dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}=1\)

Nên \(LINH< 1\left(đpcm\right)\)

Bình luận (0)
Xem chi tiết
dream XD
Xem chi tiết
Lê Thị Thục Hiền
2 tháng 7 2021 lúc 8:55

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\) 

Bình luận (0)
Trương Minh Quang
10 tháng 10 2022 lúc 18:49

CS AI XEM S** KO

Bình luận (0)
ANH HOÀNG
Xem chi tiết
Lấp La Lấp Lánh
30 tháng 9 2021 lúc 21:37

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\)

Bình luận (0)
Sir Nghi
Xem chi tiết
Hang Nguyen
Xem chi tiết
Akai Haruma
24 tháng 7 2021 lúc 0:09

Bạn tham khảo lời giải tại đây:

https://olm.vn/hoi-dap/detail/81621153379.html

Bình luận (0)
Linh Lưu
Xem chi tiết
 Mashiro Shiina
15 tháng 8 2017 lúc 21:33

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\left(\dfrac{1}{1^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^2}-\dfrac{1}{4^2}\right)+...+\left(\dfrac{1}{9^2}-\dfrac{1}{100^2}\right)\)

\(=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

\(=\dfrac{1}{1}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

Bình luận (0)
Nguyễn Huy Tú
15 tháng 8 2017 lúc 21:38

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Bình luận (0)
Ninh Nguyễn
Xem chi tiết
Dương Nguyễn
1 tháng 6 2017 lúc 11:29

Ta có:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)

= \(1-\dfrac{1}{10^2}\)

\(1-\dfrac{1}{10^2}< 1\) nên:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).

Bình luận (0)
Lê Phương Linh
Xem chi tiết

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

Bình luận (0)
Nguyễn Ngọc Anh Minh
10 tháng 10 2023 lúc 14:11

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/

\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)

\(=1-\dfrac{1}{10^2}< 1\)

 

Bình luận (0)
Kiều Vũ Linh
10 tháng 10 2023 lúc 14:13

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

Bình luận (0)