\(\left(n-1\right)^2\cdot\left(n+1\right)+\left(n^2-1\right)\)

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

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

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

Vì n; n-1; n+1 là 3 số nguyên liên tiếp

=> \(n\left(n-1\right)\left(n+1\right)⋮3\)         (1)

Vì n; n-1 là 2 số nguyên liên tiếp

\(\Rightarrow n\left(n-1\right)⋮2\)

\(\Rightarrow n\left(n-1\right)\left(n+1\right)⋮2\)        (2)

Từ (1) và (2)

=>\(n\left(n-1\right)\left(n+1\right)⋮6\)

Hay \(\left(n-1\right)^2\cdot\left(n+1\right)+\left(n^2-1\right)⋮6\) 

Vậy....

\(\left(a^2+a\right)^2+3\left(a^2+a\right)-10\)

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

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

Câu 1 : 

a, \(x^3-9x=x\left(x^2-9\right)=x\left(x-3\right)\left(x+3\right)\)

b, \(x^2-25+y^2-2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)

c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)

d, \(\left(x+2\right)^2-x\left(x+3\right)=10\)

\(\Leftrightarrow x^2+4x+4-x^2-3x-10=0\)

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

Câu 2 :

a, \(\left(x^2-4y^2\right):\left(x+2y\right)=\left(x-2y\right)\left(x+2y\right):\left(x+2y\right)=x-2y\)

b, \(\frac{-2x}{x-5}-\frac{10}{5-x}=\frac{-2x}{x-5}+\frac{10}{x-5}=\frac{-2x+10}{x-5}=\frac{-2\left(x-5\right)}{x-5}=-2\)

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

a, Xét tứ giác ABDC có: AM=MD (gt) ; BM=MC (gt)

=> ABDC là hình bình hành

b,Để ABDC là hình thoi => AB = AC => \(\Delta ABC\)cân

c, I đâu ra vậy bạn?

\(a^2-2a+1-b^2\)

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

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