???

Câu 1: 

\(a^2+b^2-a^2b^2+ab-a-b\)

\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)

\(=-a^2\left(b-1\right)\left(b+1\right)+\left(b-1\right)\left(a+b\right)\)

\(=\left(b-1\right)\left(-a^2b-a^2+a+b\right)\)

\(=\left(b-1\right)\cdot\left[-b\left(a^2-1\right)-a\left(a-1\right)\right]\)

\(=\left(b-1\right)\left(a-1\right)\left[-b\left(a+1\right)-a\right]\)

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

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

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

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

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

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

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

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

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

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)

\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)

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

\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)

\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)

Mà \(x\ne y\) nên \(xy+xz+yz-xyz\left(x+y+z\right)=0\)

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

Đpcm

Từ gt ta có : (x² - yz)y(1 - yz) = (y² - xz)x(1 - yz)

=> 0 = VT - VP = (x²y - x³yz - y²z - xy²z²) - (xy² - xy³z  - x²z - x²yz²) = xy(x - y) - xyz(x² - y²) + z(x² - y²) + xyz²(y - x)

= (x - y)[xy - xyz(x + y) + z(x + y) - xyz²] = (x - y)(xy + yz + xz - xyz(x + y + z)]

Vì\(x\ne y\Rightarrow x-y\ne0\) nên xy + yz + xz - xyz(x + y + z) = 0 => xy + yz + xz = xyz(x + y + z)

Bạn ko hiểu chỗ nào thì hỏi mình nhé!

ngu quá có thế cũng không làm được

Nguyễn Minh Phương trẻ trâu quá giỏi làm đi ko làm đc thì câm ko làm đc mà  oai thì ăn chửi

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)

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

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

\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)

\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)

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

\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)

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

Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)