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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)
t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !
=(x-y)(z-x)(z-y)(x+y+z)
\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)
\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)
\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)
Mk ko chắc
Phân tích các đa thức sau thành nhân tử
a) (x+y+z)^3 - x^3 - y^3 - z^3
b) x^4 + 2012x^2 + 2011x + 2012
= x3 + y3 + z3 + 3x2yz + 3xy2z + 3xyz2 - x3 -y3 - z3
=3x2yz + 3xy2z + 3xyz2
= 3xyz( x + y + z)
b.
x^4+2012x^2+2012x-x+2012=
(x^4-x)+2012(x^2+x+1)=
x(x-1)(x^2+x+1)+2012(x^2+x+1)=
(x+2012)(x^2+x+1)
bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
Ta có: (x-y)^3+(y-z)^3+(z-x)^3
Bạn để ý thấy (x-y)^3+(y-z)^3 là hằng đẳng thức dạng A^3+B^3=(A+B)(A^2-AB+B^2). Vậy ta có thể phân tích (x-y)^3+(y-z)^3 như sau
(x-y+y-z)((x-y)^2-(x-y)(y-z)+(y-z)^2)
(x-z)((x-y)^2-(x-y)(y-z)+(y-z)^2)
-(z-x)((x-y)^2-(x-y)(y-z)+(y-z)^2)
cách khác:
Đặt: \(x-y=a;\)\(y-z=b;\)\(z-x=c\)
suy ra: \(a+b+c=0\)
=> \(a+b=-c\)
=> \(\left(a+b\right)^3=-c^3\)
=> \(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\)
<=> \(a^3+b^3+c^3=-3ab\left(a+b\right)\)
<=> \(a^3+b^3+c^3=-3ab\left(-c\right)=3abc\)
Thay trở lại đc: \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) (x - 1)(x + l)(x - 2)(x - 4). b) (x - 2)( x 2 + 4).
c) 2y(3 x 2 + y 2 ). d) 2(x + y + z) ( a - b ) 2 .
a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)
\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)
b. \(x^3-2x^2+4x-8\)
\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)
\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
c. \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)
\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)
\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)
\(=2\left(a-b\right)^2\left(x+y+z\right)\)