Lời giải:

$S=1-3+3^2-3^3+...-3^{2021}+3^{2022}$

$3S=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}$

$\Rightarrow S+3S=3^{2023}-1$

$\Rightarrow 4S=3^{2023}-1$

$\Rightarrow 4S-3^{2023}=-1$

Ta có S = 1 + 3 + 3² + ... + 3²⁰²²

          3S = 3 + 3² + 3³ + ... + 3²⁰²³

          2S = (  3 + 3² + 3³ + ... + 3²⁰²³ ) - ( 1 + 3 + 3² + ... + 3²⁰²² )

                = 3²⁰²³ - 1

⇒ 4S - 2²⁰²³ = 2( 3²⁰²³ - 1 ) - 2²⁰²³ 

                     = 2 . 3²⁰²³ - 2 - 3²⁰²³

                     = 3²⁰²³( 2 - 1 ) - 2

                     = 3²⁰²³ - 2

Vậy 4S = 3²⁰²³ - 2

A=(1-2)+(3-4)+...+(2021-2022)+2023

=2023-(1+1+1+...+1)

=2023-1011

=1012

222222222222222222222222222222222222222222222222222222222222

\(4A-3^{2023}\) hay \(4A=3^{2023}\) hả bạn

\(A=1-3+3^2-3^3+...+3^{2021}-3^{2022}\)

\(3A=3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\)

\(3A-A=\left(1-3+3^2-3^3+...+3^{2021}-3^{2022}\right)-\left(3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\right)\)

\(2A=3^{2023}-1\)

\(\Rightarrow A=\left(3^{2023}-1\right)\div2\)

\(\text{cái này mình sợ sai nên bạn có thể nhờ cô chữa}\)

4A - 3^{2023 nha bạn}

=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023

=0+0+...+0-1-2023

=-2024

Bằng 2024

\(S=1+3+3^2+...+3^{2022}\\ 3S=3+3^2+3^3+...+3^{2023}\\ 3S-S=\left(3+3^2+3^3+...+3^{2023}\right)-\left(1+3+3^2+...+3^{2022}\right)\\ 2S=3^{2023}-1\\4S=\dfrac{3^{2023}\times2-1\times2}{2}\\ 4S=\dfrac{\left(3^{2023}-1\right)\times2}{2}\\ 4S=3^{2023}-1\\ 4S-3^{2023}=3^{2023}-1-3^{2023}\\ 4S-3^{2023}=\left(-1\right)\)

\(B=\dfrac{\dfrac{2ab}{3}-\dfrac{3ab}{2}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{\dfrac{4ab}{6}-\dfrac{9ab}{6}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{-\dfrac{5ab}{6}}{-\dfrac{5bb}{6}}=\dfrac{ab.\dfrac{5}{6}}{bb.\dfrac{5}{6}}\)

\(=\dfrac{ab}{bb}=\dfrac{a}{b}\)

Với \(a=\dfrac{2021}{2022};b=\dfrac{2023}{2022}\), ta được:

\(B=\dfrac{2021}{2022}:\dfrac{2023}{2022}=\dfrac{2021}{2022}.\dfrac{2022}{2023}=\dfrac{2021}{2023}\)

\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)

Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`

Câu 2: A

Câu 3: B

Câu 4: D

Câu 5: A

Câu 6: D

1c

2a

3b

4c

5a

6d

1. C                            2. A

3. B                            4. C

5. A                             6. D