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

b.

thêm bớt 16x^2y^2

\(64x^4+y^4=64x^4+16x^2y^2+y^4-16x^2y^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)

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

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

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

hình như bạn sai đề rồi mình cũng không giải được

bạn kiểm tra lại đề đi nhé 

(thanks)

Phát ơi ,đề sai rồi , bạn kiểm tra lại nhé :(

1) \(x^2-2x+1+x^2y-xy=\left(x-1\right)^2+xy\left(x-1\right)=\left(x-1\right)\left(x+xy-1\right)\)

2) \(x^2+6x+9+x^2y+3xy\)

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

\(=\left(x+3\right)\left(x+xy+3\right)\)

\(x^3+xy^2-9xz^2-2xyz\)

\(=x\left(x^2+y^2-9z^2-2yz\right)\)

1) x²-2xy+y²-x+y
(=) (x-y)²-(x-y)
(=) [(x-y)-1].(x-y)
(=) (x-y-1).(x-y)
C= (x-y)(x²+xy+y²)-x(x²-y)+y(y²-x)
(=) x³-y³-x³+xy+y³-xy

(=)(x³-x³)+(-y³+y³)+(xy-xy)
(=) 0

đúng nha

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

1) \(x^2.y^2-1\)

\(=\left(x.y\right)^2-1^2\)

\(=\left(x.y-1\right).\left(x.y+1\right)\)

Chúc bạn học tốt!

Làm cho mk đi @Ender Dragon Boy Vcl

(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)

=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)

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

=(x−y+z)x+(z−y)(z−y−x+y−z)=(x−y+z)x+(z−y)(z−y−x+y−z)

=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)

=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy

=x2

đc chưa

đánh chữ mỏi tay :V

\(x;y;z\rightarrow q;h;p\)

\(=\left(q^2+h^2+p^2\right)\left(q^2+h^2+p^2+2qh+2hp+2qp\right)+\left(qh+hp+pq\right)^2\)

\(Dat:\hept{\begin{cases}q^2+h^2+p^2=f\\qh+hp+qp=g\end{cases}}\Rightarrow\left(p^2+h^2+q^2\right)\left(p+q+h\right)^2+\left(qh+pq+ph\right)^2\)

\(=f\left(f+2g\right)+g^2=f^2+2fg+g^2=\left(f+g\right)^2=\left(q^2+h^2+p^2+qh+hp+pq\right)^2\)

shitbo Cho đệ sửa lại bài SP chứ bài SP dài quá ạ:p

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

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

Đặt \(x^2+y^2+z^2=a;xy+yz+zx=b\)

\(\Rightarrow a\left(a+2b\right)+b^2=a^2+2ab+b^2=\left(a+b\right)^2=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

Đặt \(x^4+y^4+z^4=a;x^2+y^2+z^2=b;x+y+z=c\)

Ta có:\(2a-b^2-2bc^2+c^4\)

\(=2a-2b^2+b^2-2bc^2+c^4\)

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

Lại có:

\(a-b^2=-2\left(x^2y^2+y^2z^2+z^2x^2\right);b-c^2=-2\left(xy+yz+zx\right)\)( Nhác quá hơi tắt xíu )

Thay vào ta được:

\(2\left(a-b^2\right)+\left(b-c^2\right)^2\)

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

\(=4xyz\left(x+y+z\right)\)