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Ta có \(xy+xz+yz=xyz\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z=\frac{xy+xz+yz}{xyz}\left(1\right)\)
Ta lại có \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x\left(1-yz\right)-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)
Vậy ta có đpcm
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)
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
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 \(yz+xy+xz=xyz\left(x+y+z\right)\)
\(VP=\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=x^2+2xy+2xz+y^2+2yz+z^2-x^2-y^2-z^2\)
\(=2xy+2yz+2xz=2\left(xy+yz+xz\right)=VP\)
Suy ra điều phải chứng minh
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-yz\right)}\)
\(\Rightarrow\left(x^2-yz\right)y\left(1-yz\right)=\left(y^2-xz\right)x\left(1-yz\right)\)
\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2=xy^2-x^2z-xy^3z+x^2yz^2\)
\(\Rightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+x^2z+xy^3z-x^2yz^2=0\)
\(\Rightarrow xy\left(x-y\right)-xyz\left(x-y\right)\left(x+y+z\right)+z\left(x-y\right)\left(x+y\right)=0\)
\(\Rightarrow\left(x-y\right)\left[xy-xyz\left(x+y+z\right)+xz+yz\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\xy+yz+zx=0\end{cases}}\)
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)
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