Ta có:

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

\(=2n^2-3n-2n^2-2n\)

\(=-5n\)chia hết cho 5.

Vậy \(n\left(2n-3\right)-2n\left(n+1\right)\)chia hết cho 5.

Ta có:

\(x^5+x^4+1\)

\(=x^5+x^4+x^3+1-x^3\)

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

\(=x^3\left(x^2+x+1\right)+\left(1-x\right)\left(1+x+x^2\right)\)

\(=\left(x^2+x+1\right)\left(x^3+1-x\right)\)

\(a^5+a^4+1=a^5+a^4+a^3-a^3+a^2-a^2+a-a+1\)

                       \(=a^5+a^4+a^3-a^3-a^2-a+a^2+a+1\)  

                       \(=\left(a^5+a^4+a^3\right)-\left(a^3+a^2+a\right)-\left(a^2+a+1\right)\)

                       \(=a^3\left(a^2+a+1\right)-a\left(a^2+a+1\right)+\left(a^2+a+1\right)\)

                        \(=\left(a^2+a+1\right)\left(a^3-a+1\right)\)

#by_Suho

\(16x^2+24xy+9y^2=\left(4x\right)^2+2.4x.3y+\left(3y\right)^2=\left(4x+3y\right)^2\)

Thay x = -1 ; y= 0

\(\left(4x+3y\right)^2=\left[4.\left(-1\right)+3.0\right]^2=16\)

Hay \(16x^2+24xy+3y^2=16\)

A = x(x - 6) + 10

A = x² - 6x + 10

A = x² - 2.3.x + 3² + 1

A = (x - 3)² + 1

Vì  (x - 3)² \(\ge\)0 với mọi x

=>  (x - 3)² + 1 \(\ge\)1 

=> A \(\ge\)1

=> A luôn dương với mọi x (Đpcm)