Gọi 3 số tự nhiên liên tiếp là

(n-1); n; (n+1). Theo đề bài

\(n\left(n+1\right)-n\left(n-1\right)=4040\)

\(\Leftrightarrow n^2+n-n^2+n=4040\Rightarrow n=2020\)

=> 3 số TN liên tiếp cần tìm là

2019; 2020; 2021

Ta có: x,y \(\in\)N nên x,y \(\ge\)0

Nếu \(x\ge2\)thì \(x^3\ge8\)và \(x^2\ge4\)nên VT > 0 ( Loại )

Nếu x = 1 thì: -1 + 4y = 0 ( Loại )

Nếu x = 0 thì 4y - 8 = 0 \(\Rightarrow\)y = 2 ( Thỏa mãn )

Vậy x=0; y = 2.

x³ - 7x² - 21x + 27

= ( x³ + 27 ) - ( 7x² + 21x )

= ( x + 3 )( x² - 3x + 9 ) - 7x( x + 3 )

= ( x + 3 )( x² - 3x + 9 - 7x )

= ( x + 3 )( x² - 10x + 9 )

= ( x + 3 )( x² - x - 9x + 9 )

= ( x + 3 )[ x( x - 1 ) - 9( x - 1 ) ]

= ( x + 3 )( x - 1 )( x - 9 )

\(xy+2y=x+4\Leftrightarrow x\left(y-1\right)=4-2y\Leftrightarrow x=\frac{4-2y}{y-1}\left(y\ne1\right)\) 

\(\Leftrightarrow x=\frac{2+2-2y}{y-1}=\frac{2-2\left(y-1\right)}{y-1}=\frac{2}{y-1}-2\)

Để x nguyên \(\Rightarrow\frac{2}{y-1}\) nguyên hay y-1 là ước của 2

\(\Rightarrow y-1=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow y=\left\{-1;0;2;3\right\}\Rightarrow x=\left\{-3;-4;0;-1\right\}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(\left(x^2+4x+x+4\right)\left(x^2+2x+3x+6\right)-24\)

\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+4=a\) ta có

\(a.\left(a+2\right)-24\)

\(a^2+2a-24\)

\(a^2+6a-4a-24\)

\(a\left(a+6\right)-4\left(a+6\right)\)

\(\left(a+6\right)\left(a-4\right)\)

\(\left(x^2+5x+4+6\right)\left(x^2+5x+4-4\right)\)

\(\left(x^2+5x+10\right)\left(x^2+5x\right)\)

   \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

   Đặt \(x^2+5x+5=a\)

   Suy ra  \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

               \(=\left(a+1\right)\left(a-1\right)-24\)

                 \(=a^2-1-24=a^2-25=\left(a-5\right)\left(a+5\right)\)

  Do đó 

         \(\left(a+5\right)\left(a-5\right)=x\left(x^2+5x+10\right)\left(x+5\right)\)

               Vậy \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=x\left(x^2+5x+9\right)\left(x+5\right)\)

\(P=\left(x^2+x +1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)-x^8-x^4\)

\(P=\left[\left(x^2+1\right)^2-x^2\right]\left(x^4-x^2+1\right)-x^8-x^4\)

\(P=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)-x^8-x^4\)

\(P=\left(x^4+1\right)^2-x^4-x^8-x^4\)

\(P=1\)