题目内容
16.求方程组$\left\{\begin{array}{l}{x\sqrt{yz}+y\sqrt{xz}=39-xy}\\{y\sqrt{xz}+z\sqrt{xy}=52-yz}\\{z\sqrt{xy}+x\sqrt{yz}=78-xz}\end{array}\right.$的正数解.分析 设$\sqrt{xy}$=s,$\sqrt{yz}$=t,$\sqrt{xz}$=u,只考虑x,y,z>0d的情况.化为xy=s2,yz=t2,xz=u2,x=$\frac{su}{t}$,y=$\frac{st}{u}$,z=$\frac{tu}{s}$,于是原方程组化为:$\left\{\begin{array}{l}{su+st=39-{s}^{2}}\\{st+tu=52-{t}^{2}}\\{tu+su=78-{u}^{2}}\end{array}\right.$,可得s+t+u=13.代入上述方程组解得:t=4,u=6,s=3.进而解出.
解答 解:设$\sqrt{xy}$=s,$\sqrt{yz}$=t,$\sqrt{xz}$=u,只考虑x,y,z>0d的情况.
则xy=s2,yz=t2,xz=u2,x=$\frac{su}{t}$,y=$\frac{st}{u}$,z=$\frac{tu}{s}$,
∴原方程组化为:$\left\{\begin{array}{l}{su+st=39-{s}^{2}}\\{st+tu=52-{t}^{2}}\\{tu+su=78-{u}^{2}}\end{array}\right.$,
∴(s+t+u)2=169,解得s+t+u=13.
代入上述方程组解得:t=4,u=6,s=3.
∴$\left\{\begin{array}{l}{xy=9}\\{yz=16}\\{xz=36}\end{array}\right.$,
解得$\left\{\begin{array}{l}{x=\frac{9}{2}}\\{y=2}\\{z=8}\end{array}\right.$.
点评 本题考查了“换元法”解方程组、乘法公式的应用,考查了推理能力与计算能力,属于中档题.
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