\(M=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\)

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

    \(=\frac{\left(xy+yz+zx\right)^2-2x^2yz-2xyz^2-2x^2yz}{xyz}\)

    \(=\frac{0-2xyz\left(x+y+z\right)}{xyz}\)

    \(=0-2\left(x+y+z\right)\)

    \(=0-2.\left(-1\right)=0-\left(-2\right)=2\)

Chúc bạn học tốt.

Ta có $xy+yz+zx=xyz$

$\Leftrightarrow\dfrac1x+\dfrac1y+\dfrac1z=1.$

Đặt $a=\dfrac1x,\quad b=\dfrac1y,\quad c=\dfrac1z.$

Khi đó $a+b+c=1.$

Ta có $H=\dfrac{c}{1+9a^2}+\dfrac{a}{1+9b^2}+\dfrac{b}{1+9c^2}.$

Với $t>0$, ta có $\dfrac1{1+9t^2}\ge1-\dfrac32t$ vì $\dfrac1{1+9t^2}-\left(1-\dfrac32t\right)$

$=\dfrac{\frac32t(3t-1)^2}{1+9t^2}\ge0.$

Do đó $H\ge c\left(1-\dfrac32a\right)+a\left(1-\dfrac32b\right)+b\left(1-\dfrac32c\right)$

$=a+b+c-\dfrac32(ab+bc+ca)$

$=1-\dfrac32(ab+bc+ca).$

Mặt khác, $(a+b+c)^2\ge3(ab+bc+ca)$

$\Rightarrow ab+bc+ca\le\dfrac13.$

Suy ra $H\ge1-\dfrac32\cdot\dfrac13$$=\dfrac12.$

Dấu bằng xảy ra khi $a=b=c=\dfrac13$

$\Leftrightarrow x=y=z=3.$

Vậy $H_{\min}=\dfrac12.$

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

Vì xyz=1 nên \(x\ne0;y\ne0;z\ne0\)

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

Tương tự \(\frac{1}{1+y+yz}=\frac{xz}{\left(1+y+yz\right)xz}=\frac{xz}{xz+z+1}\)

Khi đó \(M=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{1+z+xz}=\frac{z+xz+1}{z+zx+1}=1\)

Sorry mình mới học lớp 5

mk cx vậy

Ta có:

\(xy+yz+zx=\frac{\left(x+y+z\right)^2-x^2-y^2-z^2}{2}=\frac{7^2-23}{2}=13\)

Ta lại có:

\(xy+z-6=xy+z+1-x-y-z=\left(x-1\right)\left(y-1\right)\)

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

\(=\frac{x+y+z-3}{xyz-xy-yz-zx+x+y+z-1}=-1\)