Đáp án:
$x=1,y=2,z=3$
Giải thích các bước giải:
$\begin{cases} \dfrac{xy}{x+y}=\dfrac{2}{3}\\\dfrac{yz}{y+z}=\dfrac{6}{5}\\\dfrac{xz}{x+z}=\dfrac{3}{4} \end{cases}$
$⇔$ $\begin{cases} \dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{5}{6}\\\dfrac{x+z}{xz}=\dfrac{4}{3} \end{cases}$
$⇔$ $\begin{cases} \dfrac{x}{xy}+\dfrac{y}{xy}=\dfrac{3}{2}\\\dfrac{y}{yz}+\dfrac{z}{yz}=\dfrac{5}{6}\\\dfrac{x}{xz}+\dfrac{z}{xz}=\dfrac{4}{3}\end{cases}$
$⇔$ $\begin{cases} \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}(1)\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{6}(2)\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{4}{3}(3) \end{cases}$
Lấy $(1)+(2)+(3)$ vế tương ứng ta được :
$⇔\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{3}{2}+\dfrac{6}{5}+\dfrac{4}{3}$
$⇔\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}=\dfrac{11}{3}$
$⇔\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{11}{6} (4)$
Lấy $(4)-(1)$ vế tương ứng :
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{11}{6}-\dfrac{3}{2}$
$⇔\dfrac{1}{z}=\dfrac{1}{3}$
$⇔ z=3$
Lấy $(4)-(2)$ vế tương ứng :
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{y}-\dfrac{1}{z}=\dfrac{11}{6}-\dfrac{5}{6}$
$⇔\dfrac{1}{x}=1$
$⇔ x=1$
Lấy $(4)-(3)$ vế tương ứng :
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x}-\dfrac{1}{z}=\dfrac{11}{6}-\dfrac{4}{3}$
$⇔\dfrac{1}{y}=\dfrac{1}{2}$
$⇔y=2$
Vậy $x=1,y=2,z=3$
