x=2022 

=>x+1=2023

A=x^50-x^49(x+1)+x^48(x+1)-...+x^2(x+1)-x(x+1)+x+2

=x^50-x^50-x^49+x^49+...+x^3+x^2-x^2-x+x+2

=2

Bạn nhìn tạm nha.

Xét VT : x+3x+5x+7x+......+2023x

Số hạng của dãy số trên là : \(\dfrac{2023-1}{2}+1=1012\left(sốhạng\right)\)

Tổng số  của dãy số trên là : \(\dfrac{\left(2023x+x\right).1012}{2}\text{=}1012x.1012\)

Do đó : ta có :

\(1012x.1012\text{=}2023.2024\)

\(1012x\text{=}4046\)

\(x\text{=}\dfrac{2023}{506}\)

VT = x + 3x + 5x + 7x +... + 2023x = [(2023 - 1):2 +1] . (2023+1)x = 1012. 2024x = 2048288x

VP= 2023 . 2024= 4094552

VT=VP <=> 2048288x =4094552

<=>\(x\approx2\)

2023(10-a) min khi 10-a=0

=>a=10

Có `xyz=2023=>2023=xyz` 

Thay vào ta có :

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

`1/2023xx1/5+1/2023xx8/5-1/2023xx16/20`

`=1/2023xx(1/5+8/5-16/20)`

`=1/2023xx(1/5+8/5-4/5)`

`=1/2023xx5/5=1/2023`

ĐKXĐ : \(x\ge-2\)

\(\sqrt{1+\left(x+2\right).\sqrt{1+\left(x+3\right).\left(x+5\right)}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\sqrt{x^2+8x+16}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\left(x+4\right)}=2023x+1\) (Do \(x\ge-2\Rightarrow x+4>0\))

\(\Leftrightarrow\sqrt{x^2+6x+9}=2023x+1\)

\(\Leftrightarrow x+3=2023x+1\) (Do \(x\ge-2\Rightarrow x+3>0\)

\(\Leftrightarrow x=\dfrac{1}{1011}\)(tm) 

Vậy tập nghiệm \(S=\left\{\dfrac{1}{1011}\right\}\)

\(0^{2020}\cdot1^{2021}\cdot....\cdot21^{2120}=0\cdot1^{2021}\cdot...\cdot21^{2120}=0\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)

\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)

\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)

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

x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x

2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x

2023x - 2022x = 1 - 1/2023

x = 2022/2023