\(\dfrac{1}{\left(x-y\right)\left(z^2+yz-x^2-xz\right)}=\dfrac{1}{\left(x-y\right)\left[\left(z-x\right)\left(z+x\right)+y\left(z-x\right)\right]}=\dfrac{1}{\left(z-x\right)\left(x-y\right)\left(x+y+z\right)}\)

Tương tự: \(\dfrac{1}{\left(y-z\right)\left(x^2+xz-y^2-yz\right)}=\dfrac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}\)

\(\dfrac{1}{\left(z-x\right)\left(y^2+xy-z^2-xz\right)}=\dfrac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}\)

\(\Rightarrow M=\dfrac{y-z-z+x-x+y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}\\ M=\dfrac{2}{\left(x-y\right)\left(z-x\right)\left(x+y+z\right)}\)

\(a,\) Bổ sung điều kiện: \(x\ne y\ne z\)

\(b,\) Đề bài ko thỏa mãn điều kiện nên không tính đc M

Đề thiếu kìa :vv

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\left(\dfrac{1}{z}\right)^3=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)

\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}+3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)

\(\Rightarrow\dfrac{1}{z^3}=-\dfrac{1}{x^3}-\dfrac{3}{x^2y}-\dfrac{3}{xy^2}-\dfrac{1}{y^3}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot-\dfrac{1}{z}\)

\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=3\cdot\dfrac{1}{xyz}\)

\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)

\(\Rightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)

\(\Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vậy \(A=3\)

\(x;y;z\ne0\). Giả thiết của đề bài:

\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{z+x}\Leftrightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{x+z}{xz}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{x}+\frac{1}{z}\Leftrightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}.\)

=> x = y = z

Do đó, M = 1.

Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)

Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)

=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)

Giả thiết tương đương xy + yz + zx = 0.

Từ đó dễ dàng chứng minh được \(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3xy.yz.zx=3x^2y^2z^2\Leftrightarrow\dfrac{\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3}{3x^2y^2z^2}=\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{zx}{y^2}\).

Ta có: \(x^2+y^2-z^2\)

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

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

\(=-2xy\)

Ta có: \(x^2+z^2-y^2\)

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

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

\(=-2xz\)

Ta có: \(y^2+z^2-x^2\)

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

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

\(=-2yz\)

Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)

\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)

\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)

\(=\dfrac{-3}{2}\)