Có `xyz=2023=>2023=xyz` 

Thay vào ta có :

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

TH1: \(x+y+z+t=0\)

\(P=\left(1+\dfrac{x+y}{z+t}\right)^{2023}+\left(1+\dfrac{y+z}{x+t}\right)^{2023}+\left(1+\dfrac{z+t}{x+y}\right)^{2023}+\left(1+\dfrac{t+x}{y+z}\right)^{2023}\)

\(=\left(\dfrac{x+y+z+t}{z+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+t}\right)^{2023}+\left(\dfrac{x+y+z+t}{x+y}\right)^{2023}+\left(\dfrac{x+y+z+t}{y+z}\right)^{2023}\)

\(=0+0+0+0=0\) là số nguyên (thỏa mãn)

TH2: \(x+y+z+t\ne0\), áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2023x+y+z+t}=\dfrac{y}{x+2023y+z+t}=\dfrac{z}{x+y+2023z+t}+\dfrac{t}{x+y+z+2023t}\)

\(=\dfrac{x+y+z+t}{\left(2023x+y+z+t\right)+\left(x+2023y+z+t\right)+\left(x+y+2023z+t\right)+\left(x+y+z+2023t\right)}\)

\(=\dfrac{x+y+z+t}{2026\left(x+y+z+t\right)}=\dfrac{1}{2026}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2023x+y+z+t}=\dfrac{1}{2026}\\\dfrac{y}{x+2023y+z+t}=\dfrac{1}{2026}\\\dfrac{z}{x+y+2023z+t}=\dfrac{1}{2026}\\\dfrac{t}{x+y+z+2023t}=\dfrac{1}{2026}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2026x=2023x+y+z+t\\2026y=x+2023y+z+t\\2026z=x+y+2023z+t\\2026t=x+y+z+2023t\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4x=x+y+z+t\\4y=x+y+z+t\\4z=x+y+z+t\\4t=x+y+z+t\end{matrix}\right.\)

\(\Rightarrow4x=4y=4z=4t\) (vì đều bằng \(x+y+z+t\))

\(\Rightarrow x=y=z=t\)

Do đó:

\(P=\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}+\left(1+\dfrac{x+x}{x+x}\right)^{2023}\)

\(=2^{2023}+2^{2023}+2^{2023}+2^{2023}\)

\(=4.2^{2023}=2^{2025}\in Z\)

Em kiểm tra lại đề, 2 ngoặc cuối bị giống nhau, chắc em ghi nhầm

Chứng minh biểu thức thế nào em?

ta có biểu thức:

x⁵ − 2023x⁴ − 2023x³ − 2023x² − 2023x − 2010

thay x = 2024.

cách tính hợp lí: đặt 2024 = 2023 + 1

gọi a = 2023

-> x = a + 1

thay vào:
p(x) = (a + 1)⁵ − a(a + 1)⁴ − a(a + 1)³ − a(a + 1)² − a(a + 1) − 2010

thay a = 2023, ta được kết quả là 14

vậy p(2024) = 14

Khi x=2024 nên x-1=2023

\(x^5-2023x^4-2023x^3-2023x^2-2023x-2010\)

\(=x^5-x^4\left(x-1\right)-x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)-2010\)

\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-2010\)

=x-2010

=2024-2010

=14

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 \(S=\left\{\dfrac{1}{1011}\right\}\)

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

<=>\(x\approx2\)

Haha