x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023

x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023

x-(1-1/2023)=-2024/2023

x-2022/2023=-2024/2023

x = -2024/2023+2022/2023

x = -2/2023

Vậy x = -2/2023

:(((

a) 5.3²⁰²³ = 50.3²⁰²³ - 5.9ˣ

5.9ˣ = 50.3²⁰²³ - 5.3²⁰²³

5.(3²)ˣ = 5.3²⁰²³.(10 - 1)

5.(3²)ˣ = 5.3²⁰²³.9

3²ˣ = 3²⁰²³.3²

3²ˣ = 3²⁰²⁵

2x = 2025

x = 2025/2

b) 2.3ˣ + 5.3ˣ⁺¹ = 153

3ˣ.(2 + 5.3) = 153

3ˣ.17 = 153

3ˣ = 153/17

3ˣ = 9

3ˣ = 3²

x = 2

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2023}=\dfrac{1}{a+b+c}\)

\(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

\(\left(a+b\right)\left[\dfrac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right]=0\)

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Đến đây bạn thay vào nữa là được nhé

cảm ơn nhìu

giúp em với

Áp dụng tính chất : Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) ( a; b; n ϵ N , b; n ≠ 0 )

Ta có \(\dfrac{2023^{31}+5}{2023^{32}+5}< 1\)

⇒ \(B=\dfrac{2023^{31}+5}{2023^{32}+5}< \dfrac{2023^{31}+5+2018}{2023^{32}+5+2018}=\dfrac{2023^{31}+2023}{2023^{32}+2023}=\dfrac{2023\left(2023^{30}+1\right)}{2023\left(2023^{31}+1\right)}=\dfrac{2023^{30}+1}{2023^{31}+1}=A\)Vậy A > B

Ta có 2023A = \(\dfrac{2023.\left(2023^{30}+5\right)}{2023^{31}+5}=\dfrac{2023^{31}+5.2023}{2023^{31}+5}\)

\(=1+\dfrac{2022.5}{2023^{31}+5}\)

Lại có 2023B = \(\dfrac{2023.\left(2023^{31}+5\right)}{2023^{32}+5}=\dfrac{2023^{32}+2023.5}{2023^{32}+5}\)

\(=1+\dfrac{2022.5}{2023^{32}+5}\)

Dễ thấy 2023³¹ + 5 < 2023³² + 5

\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)

\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)

\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)

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

\(2023A=\dfrac{2023^{31}+4046}{2023^{31}+2}=1+\dfrac{4044}{2023^{31}+2}\)

\(2023B=\dfrac{2023^{32}+4046}{2023^{32}+2}=1+\dfrac{4044}{2023^{32}+2}\)

mà 2023^31+2<2023^32+2

nên A>B

2030 × 4 +2023 × 2 + 3 × 2023

 =8120 + 4046 + 6069

=18235

= 4x2023+2023x2+2023 x1 + 3x2023
=2023x (4+2+3+1)
= 2023 x 10
= 20230
cảm ơn bạn đã đọc!