Đáp án:
$e)
(x^2-1)(x^7-x^4-x^3+1)\\
d) 3(x+y)(y+z)(x+z)\\
g)
24abc\\
h) (x+y+z)\left [ (x+y+z)^2-3z(x+y) -3xy\right ]\\$
Giải thích các bước giải:
$e)
x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\
=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)\\
=x^7(x^2-1)-x^4(x^2-1)-x^3(x^2-1)+(x^2-1)\\
=(x^2-1)(x^7-x^4-x^3+1)\\
d) (x+y+z)^3-x^3-y^3-z^3\\
=(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3-x^3-y^3-z^3\\
=x^3+3x^2y+3xy^2+y^3+3.(x^2+2xy+y^2)z+3xz^2+3yz^2+z^3-x^3-y^3-z^3\\
=x^3+3x^2y+3xy^2+y^3+3x^2z+6xyz+3y^2z+3xz^2+3yz^2+z^3-x^3-y^3-z^3\\
=(x^3-x^3)+(y^3-y^3)+(z^3-z^3)+(3x^2y+3x^2z)+(3xy^2+3xyz)+(3y^2z+3xyz)+(3xz^2+3yz^2)\\
=3x^2(y+z)+3xy(y+z)+3yz(y+x)+3z^2(x+y)\\
=3x(y+z)(x+y)+3z(x+y)(y+z)\\
=3(x+y)(y+z)(x+z)$
g)
Đặt $x=a+b-c,y=b+c-a,z=c+a-b$
$\Rightarrow x+y+z=a+b-c+b+c-a+c+a-b=a+b+c$
Theo câu f ta có
$(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(z+x)\\
=(a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3\\
=3(a+b-c+b+c-a)(b+c-a+c+a-b)(c+a-b+a+b-c)\\
=3.2b.2c.2a\\
=24abc\\
h) x^3+y^3+z^3-3xyz\\
=(x+y)^3-3xy(x+y)+z^3-3xyz\\
=\left [ (x+y)^3+z^3 \right ]-3xy(x+y+z)\\
=(x+y+z)^3-3z(x+y)(x+y+z)-3xy(x+y+z)\\
=(x+y+z)\left [ (x+y+z)^2-3z(x+y) -3xy\right ]\\$
