b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )

Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)

Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)

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

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

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

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

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

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

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

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

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

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

Ta có

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

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

\(\Rightarrow C>D\)

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

\(A=\dfrac{10^{2024}+1}{10^{2023}+1}=\dfrac{10\left(10^{2023}+1\right)}{10^{2023}+1}-\dfrac{9}{10^{2023}+1}=1-\dfrac{9}{10^{2023}+1}\)

\(B=\dfrac{10^{2023}+1}{10^{2022}+1}=\dfrac{10\left(10^{2022}+1\right)}{10^{2022}+1}-\dfrac{9}{10^{2022}+1}=1-\dfrac{9}{10^{2022}+1}\)

Vì \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)

\(\Rightarrow A>B\)

nhầm rồi bạn hiếu ơi

A>B

bạn có thể giải chi tiết được không ạ?

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.