n(n^3+2n^2-n-2)

n(n-2)(n-1)(n+1)

\(n^4+2n^3-n^2-2n\)

\(=n\left(n^3+2n^2-n-2\right)\)

\(=n\left[n^2\left(n+2\right)-\left(n+2\right)\right]\)

\(=n\left[\left(n^2-1\right)\left(n+2\right)\right]\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

Tích 4 số liên tiếp chia hết cho 4 nên \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮4\)

Dễ c/m \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮2\)

và \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮3\)

Mà (2,3,4) = 1 nên \(n^4+2n^3-n^2-2n⋮24\left(đpcm\right)\)

\(n^2+12\)là số chính phương nên \(n^2+12=a^2\)

\(\Leftrightarrow a^2-n^2=12\)

\(\Leftrightarrow\left(a+n\right)\left(a-n\right)=12\)

Đến đây lập bảng giá trị

đặt n^2+12=k^2

(k-n)(k+n)=12

\(\left(x+y\right)^2+3.\left(x+y\right)+2\)

\(=\left(x+y\right)^2+2.\frac{3}{2}.\left(x+y\right)+2\)

\(=\left(x+y\right)^2+2.\frac{3}{2}.\left(x+y\right)+2+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

\(=\left(x+y\right)^2+2.\frac{3}{2}.\left(x+y\right)+\left(\frac{3}{2}\right)^2-\frac{1}{4}\)

\(=\left(x+y+\frac{3}{2}\right)^2-\frac{1}{4}\)

\(=\left(x+y+\frac{3}{2}+\frac{1}{2}\right)\left(x+y+\frac{3}{2}-\frac{1}{2}\right)\)

\(=\left(x+y+2\right)\left(x+y+1\right)\)

P/s Tham khảo nha e<3

\(\left(x+y\right)^2+3\left(x+y\right)+2\)

\(=x^2+2xy+y^2+3x+3y+2\)

\(a^5-a\)

\(=a\left(a^4-1\right)\)

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

\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)\)

\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)

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

Tích 5 số nguyên liên tiếp chia hết cho 5 nên \(a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)⋮5\)

và \(5a\left(a-1\right)\left(a+1\right)⋮5\)

Suy ra \(a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)⋮5\)

Vậy \(a^5-a⋮5\left(đpcm\right)\)

\(A=n^4+2n^3-n^2-2n\)

\(=n^3\left(n+2\right)-n\left(n-2\right)\)

\(=n\left(n+1\right)\left(n-1\right)\left(n-2\right)\)

Mà \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮4\forall n\in Z\)

=> đpcm

rồi đề bài đâu

ĐỀ ĐÂU

x²+3^y =35 <=> x² = 35 -3^y => x² là số chẵn

đặt x =2k (k \(\in N\)) => (2k)² =35 -3^y \(\le35-3^0\)=34 => k² \(\le\frac{34}{4}\approx8\)=> k \(\le\sqrt{8}\approx2\)

k=0 => x=0 => 3^y =35 (vô nghiệm)

k=1 => x=2 => 3^y =35-2² =31 (vô nghiệm)

k=2 => x=4 =>3^y =35-4² = 19 (vô nghiệm)

vậy k có x;y thỏa mãn