Giải thích các bước giải:
Ta có:
$\dfrac{xy+xz}2=\dfrac{yx+yz}{3}=\dfrac{zx+zy}4=\dfrac{(xy+xz)+(yx+yz)+(zx+zy)}{2+3+4}$
$\to\dfrac{xy+xz}2=\dfrac{yx+yz}{3}=\dfrac{zx+zy}4=\dfrac{2(xy+yz+zx)}{9}$
$\to\dfrac{xy+xz}2=\dfrac{yx+yz}{3}=\dfrac{zx+zy}4=\dfrac{xy+yz+zx}{4.5}$
$\to \dfrac{xy+yz+zx-(xy+xz)}{4.5-2}=\dfrac{xy+yz+zx-(yx+yz)}{4.5-3}=\dfrac{xy+yz+zx-(zx+zy)}{4.5-4}$
$\to \dfrac{yz}{2.5}=\dfrac{zx}{1.5}=\dfrac{xy}{0.5}$
$\to \dfrac{xyz}{2.5x}=\dfrac{zxy}{1.5y}=\dfrac{xyz}{0.5z}$
$\to 2.5x=1.5y=0.5z$
$\to\dfrac{2.5x}{7.5}=\dfrac{1.5y}{7.5}=\dfrac{0.5z}{7.5}$
$\to\dfrac{x}3=\dfrac{y}5=\dfrac{z}{15}$
$\to đpcm$
