Bài 4: Cho $x,y,z\ne 0$và $x+y+z=xyz$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}$. Tính $M=\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}$

Giải

Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}$

$\Leftrightarrow {{\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)}^{2}}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+2\left( \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \right)=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+\frac{2(x+y+z)}{xyz}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+\frac{2xyz}{xyz}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+2=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=1$

$x+y+z=xyz\Rightarrow \frac{x+y+z}{xyz}=1\Rightarrow \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=1$
$\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}={{\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)}^{2}}-\left( \frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx} \right)=3-2=1$
Vậy $M$ = 1.