A = \(\dfrac{1}{2021.2022}\) + \(\dfrac{1}{2022.2023}\) + \(\dfrac{1}{2023.2024}\) + \(\dfrac{1}{2024.2025}\) - \(\dfrac{4}{2021.2025}\)

A = \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\) + \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\) + \(\dfrac{1}{2023}\) - \(\dfrac{1}{2024}\) + \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\) - \(\dfrac{1}{2021}\) + \(\dfrac{1}{2025}\)

A = (\(\dfrac{1}{2021}\) - \(\dfrac{1}{2021}\))  + (\(\dfrac{1}{2022}\) - \(\dfrac{1}{2022}\)) + (\(\dfrac{1}{2023}\) - \(\dfrac{1}{2023}\)) + (\(\dfrac{1}{2024}\) - \(\dfrac{1}{2024}\)) + (\(\dfrac{1}{2025}\) - \(\dfrac{1}{2025}\))

A = 0 + 0  +0  + 0+ ... + 0

A = 0

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)

Giúp mình vs ạ

A = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + ... - 2023 + 2024 + 2025

Xét dãy số: 1; 2; 3; 4;..; 2025 là dãy số cách đều với khoảng cách là:

                   2  - 1  = 1

Số số hạng của dãy số trên là: ( 2025 - 1) : 1  + 1 = 2025

                  Vì 2025 : 4 = 506 dư 1 

Nhóm 4 số hạng liên tiếp của A vào nhau thì được A là tổng của 506 nhóm và 2025 khi đó

A =(1-2-3+4)+(5 - 6 - 7 + 8) +...+(2021-2022-2023+2024) + 2025

A = 0 + 0 +...+ 0 + 2025

A = 2025

\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)

P=[(1-2)+(-3+4)+(5-6)+(-7+8)+...+(993-994)+(-995+996)]+997

P=[(-1)+1+(-1)+1+...+(-1)+1+(-1)+1]+997

P= 0 +0 +...+ 0 +997

P=997

1.     Giải:

Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)

 \(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)

 \(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)

Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.

⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)

Ta có bảng:

Vậy xϵ\(\left\{0;1;3;10\right\}.\)

2. Giải:

Do (2x-18).(3x+12)=0.

⇒ 2x-18=0             hoặc             3x+12=0.

⇒ 2x     =18                               3x       =-12.

⇒   x     =9                                   x       =-4.

Vậy xϵ\(\left\{-4;9\right\}.\)

3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.

S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.

S= 0 + 0 + ... + 0 + 2025.

⇒S= 2025.

Sửa đề: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

Ta có: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(3a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}\)

=>\(3a+a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

Đặt \(b=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(3b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}\)

=>\(3b+b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)

=>\(4b=-1-\frac{1}{3^{2023}}=\frac{-3^{2023}-1}{3^{2023}}\)

=>\(b=\frac{-3^{2023}-1}{4\cdot3^{2023}}\)

Ta có: \(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)

=>\(4a=1+\frac{-3^{2023}-1}{4\cdot3^{2023}}-\frac{2024}{3^{2024}}=1+\frac{-3^{2024}-3}{4\cdot3^{2024}}-\frac{8096}{4\cdot3^{2024}}\)

=>\(4a=1-\frac{3^{2024}+8099}{4\cdot3^{2024}}=1-\frac14-\frac{8099}{4\cdot3^{2024}}=\frac34-\frac{8099}{4\cdot3^{2024}}\)

=>\(4a<\frac34\)

=>\(a<\frac{3}{16}\)

mà \(\frac{3}{16}<1<\frac{20}{3}\)

nên \(a<\frac{20}{3}\)

bài làm

Ta có:

\(\frac{2024}{2023^{2} + k} = \frac{2023^{2} + 2023}{2023^{2} + k} = 1 + \frac{2023 - k}{2023^{2} + k}\)

Vậy

\(A = \sum_{k = 1}^{2023} \left(\right. 1 + \frac{2023 - k}{2023^{2} + k} \left.\right) = 2023 + \sum_{k = 1}^{2023} \frac{2023 - k}{2023^{2} + k}\)

Vì \(\frac{2023 - k}{2023^{2} + k} > 0\) khi \(k < 2023\), và bằng 0 khi \(k = 2023\), nên

\(2023 < A < 2024\)

Suy ra A ko phải là số tự nhiên

\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)

\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)

\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)

\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)

Vì \(2024>2023=>2024^{2024}>2024^{2023}\)

\(=>2024^{2024}+1>2024^{2023}+1\)

\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)

\(=>A< B\)

\(#PaooNqoccc\)

dễ

a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)

\(\dfrac{154}{155}>\dfrac{154}{155+156}\)

\(\dfrac{155}{156}>\dfrac{155}{155+156}\)

=>154/155+155/156>(154+155)/(155+156)

=>A>B

b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)

2021/2022>2021/6069

2022/2023>2022/2069

2023/2024>2023/6069

=>D>C