\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right)\)

\(B=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)....\left(\dfrac{1}{2020^2}-\dfrac{2020^2}{2020^2}\right)\)

\(B=\left(\dfrac{1-2^2}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)...\left(\dfrac{1-2020^2}{2020^2}\right)\)

\(B=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}\cdot\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}....\cdot\dfrac{\left(2020-1\right)\left(2020+1\right)}{2020^2}\) 

\(B=\dfrac{-1\cdot3}{2^2}\cdot\dfrac{-2\cdot4}{3^2}\cdot\dfrac{-3\cdot5}{4^2}\cdot....\cdot\dfrac{-2019\cdot2021}{2020}\)

\(B=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-2019}{2\cdot3\cdot4\cdot....\cdot2020}\)

\(B=\dfrac{-1\cdot-1\cdot-1\cdot....\cdot-1}{1}\)

\(B=-1\) (2019 số -1) 

Mà: \(-1< \dfrac{1}{2}\)

\(\Rightarrow B< \dfrac{1}{2}\)

 \(\dfrac{1}{2^2}\); \(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 1 ⇒ 0 > \(\dfrac{1}{2^2}\) - 1 > \(\dfrac{1}{3^2}\) - 1 >..> \(\dfrac{1}{2020^2}\) - 1

Xét dãy số 2; 3; 4;...; 2020 dãy số này có số số hạng là:

        (2020 - 2):1 + 1 = 2019 (số hạng)

Vậy B là tích của 2019 số âm nên B < 0 ⇒ B < \(\dfrac{1}{2}\)

Ta có `3A=1+1/3+....+1/3^99`

`=>3A-A=1-1/3^100`

`=>2A=1-1/3^100`

`=>A=1/2-1/(2.3^100)<1/2`

Hay `A<B`

Ta có: \(3\cdot A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

Do đó: 

\(3\cdot A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)

hay \(2\cdot A=1-\dfrac{1}{3^{100}}\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right):2\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right)\cdot\dfrac{1}{2}\)

\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{100}}< \dfrac{1}{2}\)

hay A<B

⇔A=12−12⋅3100<12⇔A=12−12⋅3100<12

hay A<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 ) )

Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)

=10

Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)

\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)

\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)

\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)

Cái này đâu có b đâu bạn

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

1/3²< 1/2.3

1/4²< 1/3.4

...

1/100²< 1/99.100

=> 1/2² + 1/3² + 1/4² + ... + 1/100²< 1/2² + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4

=> A < 3/4

Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2022}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2021}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{2022}}$

$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{2022}}$

Xét hiệu:
$A-B=\frac{1}{2}-\frac{1}{2.3^{2022}}-(1-\frac{1}{3^{2021}})$

$=\frac{1}{3^{2021}}-\frac{1}{2.3^{2022}}-\frac{1}{2}$

$=\frac{5}{2.3^{2022}}-\frac{1}{2}$

$< \frac{1}{2}-\frac{1}{2}=0$

$\Rightarrow A< B$

`A = 1/3 +1/3^2 +1/3^3 +...+1/3^2022`

`<=> 3A = 1 +1/3 +1/3^2 +...+ 1/3^2021`

`=>2A =3A-A =1+1/3 +1/3^2 +..+ 1/3^2021 - 1/3-1/3^2-1/3^3..-1/3^2022`

`2A = 1-1/3^2022`

`=> A = (1-1/3^2022) :2`

Ta thấy `1- 1/3^2022 < 1-1/3^2021`

`=> (1 -1/3^2022):2<1-1/3^2021`

Hay `A<B`