Sửa đề: Tìm x

a: \(x\left(6-x\right)^{2023}=\left(6-x\right)^{2023}\)

=>\(x\left(6-x\right)^{2023}-\left(6-x\right)^{2023}=0\)

=>\(\left(6-x\right)^{2023}\left(x-1\right)=0\)

=>\(\left[\begin{array}{l}6-x=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=1\end{array}\right.\)

b: \(5^{x}+5^{x+2}=650\)

=>\(5^{x}+5^{x}\cdot25=650\)

=>\(5^{x}\left(1+25\right)=650\)

=>\(5^{x}=\frac{650}{26}=25=5^2\)

=>x=2

c: \(2^{x+2}-2^{x}=96\)

=>\(2^{x}\cdot2^2-2^{x}=96\)

=>\(2^{x}\left(2^2-1\right)=96\)

=>\(2^{x}\cdot3=96\)

=>\(2^{x}=\frac{96}{3}=32=2^5\)

=>x=5

d: \(10^{x}:5^{y}=20^{y}\)

=>\(10^{x}=20^{y}\cdot5^{y}=100^{y}=\left(10\right)^{2y}\)

=>x=2y

Sửa đề: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

Ta có: \(\frac{x-1}{2023}+\frac{x-2}{2022}+\cdots+\frac{x-2022}{2}=2022\)

=>\(\left(\frac{x-1}{2023}-1\right)+\left(\frac{x-2}{2022}-1\right)+\cdots+\left(\frac{x-2022}{2}-1\right)=2022-2022=0\)

=>\(\frac{x-2024}{2023}+\frac{x-2024}{2022}+\cdots+\frac{x-2024}{2}=0\)

=>\(\left(x-2024\right)\left(\frac{1}{2023}+\frac{1}{2022}+\cdots+\frac12\right)=0\)

=>x-2024=0

=>x=2024

\(\)(2 - x)\(^{2023}\) = 1

\(\)(2 - x)\(^{2023}\) = 1\(^{2023}\) = (-1)\(^{2023}\)

TH1: (2 - x)\(^{2023}\) = 1\(^{2023}\)

⇒ 2 - x = 1

x = 2 - 1

x = 1

TH2: (2 - x)\(^{2023}\) = (-1)\(^{2023}\)

⇒ 2 - x = -1

x = 2 - (-1)

x = 3

Vậy x ∈ {1; 3}

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

=>(x-2023)[(x-2023)^21-1]=0

=>x-2023=0 hoặc x-2023=1

=>x=2023 hoặc x=2024

\(A=4x4x...x4\left(2023.chũ.số.4\right)\)

\(A=4^{23}=4^{20}.4^3=\overline{....6}x\overline{....4}=\overline{....4}\)

\(B=3x15x23x...x2023\)

\(B=\overline{....5}\) (trong tích có các số có tận cùng bằng 5)

đề là tìm cs tận cùng của tích ạ

Có `xyz=2023=>2023=xyz` 

Thay vào ta có :

\(\dfrac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz+1+z}{1+xz+z}=1\left(dpcm\right)\)

Lời giải:
Ta có:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2023.\frac{2024}{2023}$

$\Leftrightarrow 1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=2024$

$\Leftrightarrow 3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}=2024$

$\Leftrightarrow 3+B=2024$

$\Leftrightarrow B=2021$