Đáp án:

$\begin{array}{l}
p){x^2} - 3xy - 10{y^2}\\
 = {x^2} - 5xy + 2xy - 10{y^2}\\
 = \left( {x - 5y} \right)\left( {x + 2y} \right)\\
e){a^3} + 2{a^2} + a - a{b^2}\\
 = a.\left( {{a^2} + 2a + 1 - {b^2}} \right)\\
 = a.\left[ {{{\left( {a + 1} \right)}^2} - {b^2}} \right]\\
 = a.\left( {a + 1 + b} \right)\left( {a + 1 - b} \right)\\
s)4xy - 4{y^2} + 16 - {x^2}\\
 = 16 - \left( {{x^2} - 4xy + 4{y^2}} \right)\\
 = 16 - {\left( {x - 2y} \right)^2}\\
 = \left( {4 - x + 2y} \right)\left( {4 + x - 2y} \right)\\
t){x^2} + xy - 17x - 17y\\
 = x\left( {x + y} \right) - 17\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {x - 17} \right)\\
u){x^5} + {x^3} - {x^2} - 1\\
 = \left( {{x^2} + 1} \right)\left( {{x^3} - 1} \right)\\
 = \left( {{x^2} + 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
v)7x\left( {3x - y} \right) - 10y\left( {y - 3x} \right)\\
 = \left( {3x - y} \right)\left( {7x + 10y} \right)\\
x){x^2} + 7x - 8\\
 = {x^2} + 8x - x - 8\\
 = \left( {x + 8} \right)\left( {x - 1} \right)\\
y)5{x^2}y - 10x{y^2} = 5xy\left( {x - y} \right)\\
{\rm{w}}){\left( {3x - 4} \right)^2} - {\left( {x + 1} \right)^2}\\
 = \left( {3x - 4 + x + 1} \right)\left( {3x - 4 - x - 1} \right)\\
 = \left( {4x - 3} \right)\left( {2x - 5} \right)\\
e){x^3} + 3{x^2} + 3x - 26\\
 = {x^3} + 3{x^2} + 3x + 1 - 27\\
 = {\left( {x + 1} \right)^3} - {3^3}\\
 = \left( {x + 1 - 3} \right)\left( {{{\left( {x + 1} \right)}^2} + 3\left( {x + 1} \right) + {3^2}} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} + 2x + 1 + 3x + 3 + 9} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} + 5x + 13} \right)\\
o){x^4} + {x^3} + 2{x^2} + x + 1\\
 = {x^4} + {x^3} + {x^2} + {x^2} + x + 1\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^2} + 1} \right)\\
u){x^5} + x + 1\\
 = {x^5} - {x^2} + {x^2} + x + 1\\
 = {x^2}\left( {{x^3} - 1} \right) + {x^2} + x + 1\\
 = {x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^3} - {x^2} + 1} \right)\\
d){x^6} - {y^6}\\
 = \left( {{x^2} - {y^2}} \right)\left( {{x^4} + {x^2}{y^2} + {y^4}} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right)\left( {{x^4} + {x^2}{y^2} + {y^4}} \right)\\
a){x^3} + {y^3} + {z^3} - 3xyz\\
 = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} + {z^3}\\
 - 3{x^2}y - 3x{y^2} - 3xyz\\
 = {\left( {x + y} \right)^3} + {z^3} - 3xy\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right)\left( {{{\left( {x + y} \right)}^2} - \left( {x + y} \right).z + {z^2}} \right)\\
 - 3xy\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right)\left( {{x^2} + 2xy + {y^2} - xz - yz + {z^2} - 3xy} \right)\\
 = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xz - yz - xy} \right)
\end{array}$

Mình đã rất cố gắng để giải hết rồi , nên nếu thấy lời giải hay,dễ hiểu thì cho mình 5 sao nhé bạn !

p, x²-3xy-10y²

= x²+2xy-5xy-10y²

= (x²+2xy)-(5xy+10y²)

= x.(x+2y)-5y(x+2y)

= (x+2y).(x-5y)

r, a³+2a²+a-ab²

= a.(a²+2a+1-b²)

= a.[(a²+2a+1)-b²]

= a.[(a+1)²-b²]

= a.(a+1-b).(a+1+b)

s, 4xy-4y²+16-x²

= 16+(4xy-4y²-x²)

= 16-(4y²-4xy+x²)

= 16-(2y-x)²

= (4-2y+x).(4+2y-x)

t, x²+xy-17x-17y

= (x²+xy)-(17x+17y)

= x.(x+y)-17.(x+y)

= (x+y).(x-17)

u, x^5 + x³-x²-1

= (x^5 +x³)-(x²+1)

= x³.(x²+1)-(x²+1)

= (x²+1).(x³-1)

= (x²+1).(x-1).(x²+x+1)

v, 7x.(3x-y)-10y.(y-3x)

= 7x.(3x-y)+10y.(3x-y)

= (3x-y).(7x-10y)

x, x²+7x-8

= (x²-x)+(8x-8)

= x.(x-1)+8.(x-1)

= (x-1).(x+8)

y, 5x²y-10xy²

= 5xy.(x-2y)

w, (3x-4)²-(x+1)²

= (3x-4-x-1).(3x-4+x+1)

= (2x-5).(4x-3)

ê, x³+3x²+3x-26

= (x³-2x²)+(5x²-10x)+(13x-26)

= x².(x-2)+5x.(x-2)+13.(x-2)

= (x-2).(x²+5x+13)

ô, x^4 + x³+2x²+x+1

= (x^4 +x³+x²)+(x²+x+1)

= x².(x²+x+1)+(x²+x+1)

= (x²+1).(x²+x+1)

ơ, (x+1).(x+3).(x+5).(x+7)+1

= [(x+1).(x+7)].[(x+3).(x+5)]+1

= (x²+8x+7).(x²+8x+15)+1   (1)

Đặt x²+8x+7=y

Ta có (1)= y.(y+5)+1

= y²+5y+1

= (y²+2.y.5/2+25/4)-21/4

= (y+5/2)²-21/4

= (y+5/2-√21/2).(y+5/2+√21/2)

ư, x^5 + x+1

= x^5 +x^4 +x³-x^4 -x³-x²+x²+x+1

= (x^5 + x^4 +x³)-(x^4 +x³+x²)+(x²+x+1)

= x³(x²+x+1)-x².(x²+x+1)+(x²+x+1)

= (x²+x+1).(x³-x²+1)

đ, x^6-y^6

= (x³-y³).(x³+y³)

= (x-y).(x²+xy+y²).(x+y).(x²-xy+y²)

ă,  x³+y³+z³-3xyz

= (x³+y³)+z³-3xyz

= (x+y)³-3xy.(x+y)+z³-3xyz

= [(x+y)³+z³]-[3xy.(x+y)+3xyz]

= (x+y+z).[(x+y)²-(x+y).z+z²]-3xy.(x+y+z)

= (x+y+z).(x²+2xy+y²-xz-yz+z²)-3xy.(x+y+z)

= (x+y+z).(x²+2xy+y²-xz-yz+z²-3xy)

= (x+y+z).(x²+y²-xz-yz+z²-xy)