Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(x^4+2x^3-4x-4=\left[\left(x^2\right)^2-4\right]+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2+2+2x\right)\left(x^2-2\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)=x^2\left(x+1\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) \(x^2+y^2-x^2y^2+xy-x-y=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)=\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(=\left(1-y\right)\left[\left(x-1\right)x+y\left(x-1\right)\left(x+1\right)\right]=\left(1-y\right)\left(x-1\right)\left(x+xy+y\right)\)
c) Không phân tích được.
x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)
b: =x^3+2x^2-x^2+4
=x^2(x+2)-(x+2)(x-2)
=(x+2)(x^2-x+2)
c: =x^3-2x^2+x^2-4
=x^2(x-2)+(x-2)(x+2)
=(x-2)(x^2+x+2)
d: =(x-y)(x+y)-7(x+y)
=(x+y)(x-y-7)
a: \(=x^2\left(x-2\right)\)
b: \(=\left(x-3\right)\left(2x-9\right)\)
\(a,=x^2\left(x-2\right)\\ b,=\left(x-3\right)\left(2x-9\right)\\ c,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
b) \(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
câu a còn thiếu thì phải bạn xem lại xem
a. Đề phải là thế này:
x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)