Ta có : \(-\dfrac{3}{5}=\dfrac{-21}{35}\);\(-\dfrac{2}{7}=-\dfrac{10}{35}\)

MT : 35 

=> \(-\dfrac{3}{5}< -\dfrac{2}{7}\)

\(\dfrac{-3}{5}=\dfrac{\left(-3\right)\times7}{5\times7}=\dfrac{-21}{35}\) ; \(\dfrac{-2}{7}=\dfrac{\left(-2\right)\times5}{7\times5}=\dfrac{-10}{35}\)

⇒ \(\dfrac{-3}{5}< \dfrac{-2}{7}\)

mẫu chung =35 ta có -21/35 và -10/35

vì -21/35 < -10/35

nên -3/5<2/7

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

Bạn tham khảo lời giải tại đây:

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

M=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{17}{8^2.9^2}+\dfrac{19}{9^2.10^2}\)

=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{17}{64.81}+\dfrac{19}{81.100}\)

=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{81}+\dfrac{1}{81}-\dfrac{1}{100}\)

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

.

Mình ko biết làm bạn ạ

\(\dfrac{N}{2}=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}< 1\\ N< 2\)

Ta có: \(N=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=2\left(1-\dfrac{1}{50}\right)\)

\(=2\cdot\dfrac{49}{50}=\dfrac{49}{25}< \dfrac{50}{25}=2\)

Vậy: N<2

Ta có: =2(11⋅2+12⋅3+13⋅4+...+149⋅50)=2(11⋅2+12⋅3+13⋅4+...+149⋅50)

=2(1−150)=2(1−150)

a) \(\dfrac{2}{5}=\dfrac{4}{10}\)

\(\dfrac{4}{10}>\dfrac{3}{10}\)

b) \(\dfrac{5}{6}=\dfrac{10}{12}\)

\(\dfrac{7}{12}< \dfrac{10}{12}\)

c) \(\dfrac{1}{2}=\dfrac{2}{4}\)

\(\dfrac{3}{4}< \dfrac{2}{4}\)

d) \(\dfrac{8}{3}=\dfrac{56}{21}\)

\(\dfrac{56}{21}>\dfrac{11}{21}\)

a) \(\dfrac{3}{4}=\dfrac{3\times4}{4\times4}=\dfrac{12}{16}\)

b) \(\dfrac{1}{3}=\dfrac{1\times3}{3\times3}=\dfrac{3}{9}\)

c) \(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18}\)

a)\(\dfrac{-36}{63};\dfrac{56}{63};\dfrac{-30}{63}\)

b)\(\dfrac{110}{264};\dfrac{21}{264}\)

a) Mẫu số chung là BCNN(7, 9, 21) = 3².7 = 63

Thừa số phụ của 7 là 9, của 9 la 7, của 21 là 3. Do đó:

b) Mẫu số chung: 2³.3.11 = 264. Do đó:

\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{39}{19^2.20^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{39}{361.400}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{361}-\dfrac{1}{400}\)

\(=1-\dfrac{1}{400}< 1\)

Vậy A < 1