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9 tháng 10 2018

Sửa đề chút :

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

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

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

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

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

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

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

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

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

9 tháng 10 2018

c) x+ y3 + z3 - 3xyz

= x3 + 3x2y + 3xy2 + y3 + z3 - 3xyz - 3x2y - 3xy2

= (x+y)3 + z3  - 3xy.( z+x+y)

= (x+y+z).[(x+y)2 - (x+y).z + z2 ] - 3xy.(x+y+z)

= (x+y+z). ( x2 + 2xy + y2 - xz - yz + z2 - 3xy)

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

e) (a+b-c)2 - (a-c)2 - 2ab + 2bc

= (a+b-c - a+c).(a+b+c+a-c) - 2b.(a-c)

= b.(2a+b) - 2b.(a-c)

= b.(2a+b - a +c)

= b.( a+b+c)

xl bn nha! mk chỉ nghĩ đk 2 câu thoy, 1 câu bn kia làm r! 2 câu còn lại bn đợi người tiếp theo làm nhé

2 tháng 7 2021

a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz

= xy(X + y + z)  + yz(x + y + z) + xz(X + y + z)

= (x + y +z)(xy + yz+ xz)

b) xy(x + y) - yz(y + z) - xz(z - x)

= x2y + xy2 - y2z - yz2 - xz2 + x2z

= x2(y + z) - yz(y + z) + x(y2 - z2)

= x2(y + z) - yz(y + z) + x(y + z)(y - z)

= (y + z)(x2 - yz + xy - xz)

= (y + z)[x(x + y) - z(x + y)]

= (y + z)(x + y)(x - z)

c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)

 = x(y - z)(y + z) + yz2 - yx2 + x2z - y2z

= x(y - z)(y + z) - yz(y - z) - x2(y - z)

= (y - z)((xy + xz - yz - x2)

= (y - z)[x(y - x) - z(y - x)]

= (y - z)(x - z)(y -x) 

2 tháng 9 2018

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

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

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

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

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

1 tháng 11 2016

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)

27 tháng 9 2016

mk học lớp 7 thui

26 tháng 9 2017

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

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

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

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

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

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

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

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

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

6 tháng 7 2022

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

=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)=(xy)2+(yx)(x2+y2+3z23yz+xy3xz)

=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)=(xy)(xyx2y23z2+3yzxy+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) \left(x+y+z\right)\left(xy+yz+zx\right)-xyzb)(x+y+z)(xy+yz+zx)xyz

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

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

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

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

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

=\left(x+y\right)\left(xy+zx+zy+z^2\right)=(x+y)(xy+zx+zy+z2)

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

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

 
17 tháng 8 2017

Câu a :

\(VT=\) \(\left(x-1\right)\left(x^2+x+1\right)=x^3-1^3=VP\)

Câu b :

\(VT=\)\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4-y^4=VP\)

Tương tự bạn khai triển là ra nhé

17 tháng 8 2017

a) \(\left(x-1\right)\left(x^2+x+1\right)\)

=\(x^3+x^2+x-x^2-x-1=x^3-1\)

\(\RightarrowĐPCM\)

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

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

27 tháng 7 2017

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

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

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

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

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

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

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

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

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

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

c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)

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

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

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

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

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

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

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

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

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

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

d, \(x^3+y^3+z^3-3xyz\)

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

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

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

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