a)  \(2x^2+3x-5=2x^2-2x+5x-5\)

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

b)  \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+3z\left(x+y\right)\left(x+y+z\right)+z^3-x^3-y^3-z^3\)

\(=x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

c)  \(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)

d)  \(x^4-x^3-x^2+1=x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2-x-1\right)\)

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

c,4x²-20x+25-4x²-20x-25+40x-1

=-1

a,2(x²-2x+1)

=2(x-1)²

b,3x(x²+4x+4)

=3x(x+2)²

c,x(x²-2x+1)

x(x-1)²

d,(x³-1)-(3x²-3x)

=(x-1).(x²-x+1)-3x(x-1)

(x-1).(x²-x+1-3x)

=(x-1).(x²-4x+1)

\(B=x^2-2xy+3y^2-2x-10y+20\)

\(=x^2-2xy+y^2-2\left(x-y\right)+1+2y^2-12y+19\)

\(=\left(x-y\right)^2-2\left(x-y\right)+1+2\left(y^2-6y+9\right)+1\)

\(=\left(x-y-1\right)^2+2\left(y-3\right)^2+1\ge1\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x-y-1=0\\y-3=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=4\\y=3\end{cases}}\)

Vậy Min \(B=1\)khi \(x=4;\)\(y=3\)

\(x^{16}+x^8-2=x^{16}-x^8+2x^8-2=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)

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

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

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