a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)

\(=\left[\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\right]-\left(y^2+z^2\right)^3\)

\(=\left(x^2+y^2+z^2-x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]-\left(y^2+z^2\right)^3\)

\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4\right)-\left(y^2+z^2\right)^3\)

\(=\left(y^2+z^2\right)\left[x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-\left(y^2+z^2\right)^2\right]\)

\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-y^4-2y^2z^2-z^4\right)\)

\(=\left(y^2+z^2\right)\left(3x^4+3x^2y^2-3x^2z^2-3y^2z^2\right)\)

   = 3(y²+z²)(x⁴+x²y²-x²z²-y²z²)

   = 3(y²+z²)[x²(x²+y²)-z²(x²+y²)]

   = 3(y²+z²)(x²-z²)(x²+y²)

   = 3(y²+z²)(x-z)(x+z)(x²+y²)

b) \(\left(x+y\right)^3-x^3-y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)

\(=3x^2y+3xy^2=3xy\left(x+y\right)\)

c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2+z^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2-\left(x^3+y^3\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right).z+3z^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)

  = (x+y)[3xy+3xz+3yz+3z² ] 

  = 3(x+y)(xy+xz+yz+z²)

  = 3(x+y)[x(y+z)+z(y+z)]

  = 3(x+y)(x+z)(y+z)

a) \(\left(x^2+y^2\right)^3+\left(z^2-x^3\right)-\left(y^2+z^2\right)^3\)

\(=x^6+3x^4y^2+3x^4y^2+y^6+z^2+-x^2+-y^6+-3y^4z+-3y^2z^4+-z^6\)

\(=x^6+3x^4y^2+3x^2y^4+-3y^4z^4+-z^6+-x^2+z^2\)

b) \(\left(x+y\right)^3-x^3-y^3\)

\(=x^3+3x^2y+3xy^2+y^3+-x^3+-y^3\)

\(=\left(x^3+-x^3\right)+\left(3x^2y\right)+\left(3xy^2\right)+\left(y^3+-y^3\right)\)

\(=3x^2y+3xy^2\)

c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^2-x^3-y^3-z^3\)

\(=3x^2y+3x^2z+3xy^2+3xy^2+6xyz+3xz^2+3y^2z+3yz^2\)

P/s: Ko chắc

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

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

mk học lớp 7 thui

a. =\(\left(x^2+1\right).\left(x^2+x+1\right)\)

b = \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

c=\(\left(3z+3y\right)\left(x+y\right)\left(z-x\right)\)

d= \(3\left(ab+bc+ca\right)\)

2 câu còn lại mình ko biết

xin lỗi mình chỉ viết đc đáp án vì nó dài quá

a) \(\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]\)

\(=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)\)

\(=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)\)

Cô nghĩ phân tích đa thức này sẽ đẹp hơn:

\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)

\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)

\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]\)

\(=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)\)

\(=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

b) \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)

\(=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz\)

\(=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)\)

\(=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)\)

\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)

\(=\left(x+y\right)\left(xy+zx+zy+z^2\right)\)

\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

a) \left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3(x−y)2+(y−z)3+(z−x)3

=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]=(x−y)2+(y−z+z−x)[(y−z)2−(y−z)(z−x)+(z−x)2]

=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)=(x−y)2+(y−x)(x2+y2+3z2−3yz+xy−3xz)

=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)=(x−y)(x−y−x2−y2−3z2+3yz−xy+3xz

\left(x-y\right)^3+\left(y-z\right)^3+\left

 

=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3


 

=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\l

 

=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\

 

=\left(x-z\right)\left(

=3\left(x-y\right)\lefb) $b)(x+y+z)\left(xy+yz+zx\right)-xyz$

$=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz$

$=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz$

=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)