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Lời giải:
a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$
$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$
$=3xy(y-x)+5(y-x)=(y-x)(3xy+5)$
c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$
d.
$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
\(xy-3x-2y+6=x\left(y-3\right)-2\left(y-3\right)=\left(y-3\right)\left(x-2\right)\)
\(x^2-6xy-4z^2+9y^2=\left(x-3y\right)^2-\left(2z\right)^2=\left(x-3y-2z\right)\left(x-3y+2z\right)\)
\(b,=x^2-xy=x\left(x-y\right)\\ c,=y\left(x+1\right)+z\left(x+1\right)=\left(y+z\right)\left(x+1\right)\\ d,=x^2+ax+bx+ab\\ =x\left(x+a\right)+b\left(x+a\right)=\left(x+b\right)\left(x+a\right)\)