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

\(=\frac{\left(yz+xz+xy\right)\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)

\(=\frac{0.\left[\left(yz+xz\right)^2+xy\left(yz+xz\right)+x^2y^2\right]-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}\)

\(=\frac{-3xyz^2\left(xz+yz\right)}{x^2y^2z^2}=\frac{-3\left(xz+yz\right)}{xy}=\frac{-3.\left(-xy\right)}{xy}=3\)

Từ đề <=>\(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+zy}\Leftrightarrow xz=xy=zy\)

Có : \(zx=xy\Rightarrow y=z\left(\text{Vì }x\ne0\right),xy=zy\Rightarrow x=z\)

=> x=y=z 

tự tính M :]]

bạn nào t-i-k sai cho tớ làm lại hộ ạ :)

ai t-i-k sai thì làm lại đi :(( 

ta có xy+yz+zx=0=> \(\frac{xy+yz+zx}{xyz}=0\)

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

đặt \(\frac{1}{x}=a,\frac{1}{y}=b,\frac{1}{z}=c\Rightarrow a+b+c=0\)

ta xét \(a^3+b^3+c^3-3abc=a^3+b^3+3ab\left(a+b\right)+c^3-3ab-3abc\)

           \(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

=> \(a^3+b^3+c^3=3abc\) \(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

=> \(M=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)

=> M=3

Ta có: \(x+y+z=0\)

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

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

\(\Leftrightarrow x^2+2xy+y^2=z^2\)

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

Chứng minh tương tự ta có:

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

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

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

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

\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)

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

Vậy giá trị biểu thức là \(-\frac{3}{2}\)

Ta có \(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{x+z}\)

=> \(\frac{xyz}{xz+yz}=\frac{xyz}{xy+xz}=\frac{xyz}{xy+yz}\)

=> \(xz+yz=xy+xz=xy+yz\)(vì x ; y ;z \(\ne0\Leftrightarrow xyz\ne0\))

=> \(\hept{\begin{cases}xz+yz=xy+xz\\xy+xz=xy+yz\\xz+yz=xy+yz\end{cases}}\Rightarrow\hept{\begin{cases}yz=xy\\xz=yz\\xz=xy\end{cases}}\Rightarrow\hept{\begin{cases}z=x\\x=y\\y=z\end{cases}}\Rightarrow x=y=z\)

Khi đó M = \(\frac{x^2+y^2+z^2}{xy+yz+zx}=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1\left(\text{vì }x=y=z\right)\)

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

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

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