\(3,\left(5\right)=\dfrac{32}{9}\)

\(3,\left(5\right)=3+0,\left(5\right)=3+5.0,\left(1\right)=3+5.\dfrac{1}{9}=3+\dfrac{5}{9}=\dfrac{32}{9}\)

đáp án ....... ...¿.¿¿¿

\(\left(-39,82\right):\left(-7,24\right)\)

\(=39,82:7,24\)

\(=5,5\)

5,5 nha

a) A = 1/(1.2) + 1/(2.3) + ... + 1/[n(n + 1)]

= 1 - 1/2 + 1/2 - 1/3 + 1/n - 1/(n + 1)

= 1 - 1/(n + 1)

b) Do n ∈ ℕ

⇒ n + 1 > 0

⇒ 1/(n + 1) > 0

⇒ 1 - 1/(n + 1) < 1

Vậy A < 1

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15

c) và d) của Trí sai nên sửa lại

c) (2x - 4)/7 < 0

⇒ 2x - 4 < 0 (vì 7 > 0)

⇒ 2x < 4

⇒ x < 2

d) (5x - 8)/-10 < 0

⇒ 5x - 8 > 0 (vì -10 < 0)

⇒ 5x > 8

⇒ x > 8/5

a) \(\dfrac{x-2}{45}>0\Rightarrow x-2>0\Rightarrow x>2\)

b) \(\dfrac{x+3}{-2}>0\Rightarrow x+3< 0\Rightarrow x< -3\)

c) \(\dfrac{2x-4}{7}< 0\Rightarrow2x-4>0\Rightarrow2x>4\Rightarrow x>2\)

d) \(\dfrac{5x-8}{10}< 0\Rightarrow5x-8< 0\Rightarrow5x< 8\Rightarrow x< \dfrac{8}{5}\)

Ta có :

\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)

mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)

     \(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)

\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)

a) \(\left|a\right|=\left|b\right|\Rightarrow a=b,\forall\left|a\right|>0\left(1\right)\)

\(\left|2\right|=\left|-2\right|\Rightarrow2=-2,\left|2\right|>0\Rightarrow\left(1\right)sai\)

b) \(\left|a\right|>\left|b\right|\Rightarrow a>b,\forall\left|a\right|>b\left(1\right)\)

\(\left|-3\right|>\left|2\right|\Rightarrow-3>2,\left|-3\right|>2\Rightarrow\left(1\right)sai\)

\(\left|a\right|>a,\forall a\) (1)

|4| > 4 hay 4 > 4, vô lí, suy ra (1) sai