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a: =(x+y)^3+z^3-3xy(x+y)-3xyz
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2
=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c
=b^2(c-a)+b(c^2-a^2)+ac(c-a)
=(c-a)(b^2+ac)+b(c-a)(c+a)
=(c-a)(b^2+ac+bc+ba)
=(c-a)[b^2+bc+ac+ab]
=(c-a)[b(b+c)+a(b+c)]
=(c-a)(b+c)(b+a)
\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)
(b-c)^2+b(a-c)^2+c(a-b)^2- a^3 -b^3 -c^3 +4abc
=a[(b-c)^2-a^2)]+ b[(a-c)^2-b^2)]+c[(a-b)^2-c^2)]+4abc
=a[(b-c)^2-a^2)]+ b[(a+c)^2-b^2)]+c[(a-b)^2-c^2)]
=a(b-c-a)(b-c+a)+b(a+c-b)(a+b+c)+c(a+c...
=[-a(b-c+a)+b(a+b+c)+c(a-b-c)](a+c-b)
Bạn cứ tiếp tục phân tích cái vế trong ngoặc vuông đuọc (a+b-c)(b+c-a) là đc.
Đáp số : (a+c-b)(a+b-c)(b+c-a)
\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)
\(a,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-40\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)=\left(x^2+6x\right)\left(x^2+6x+13\right)\\ b,=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\\ =a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)-b^2c^2\left(c-a\right)+c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(a^2b^2-b^2c^2\right)-\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)-c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(a-c\right)\left[b^2\left(a+c\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(a-c\right)\left(a^2b+b^2c-b^2c+a^2c\right)\\ =a^2\left(a-b\right)\left(a-c\right)\left(b+c\right)\)
\(c,=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
