初中代数题 求解
已知p+q+r=9,p/(x²-yz)=q/(y²-zx)=r/(z²-xy)求(px+qy+rz)/(x+y+z)...
已知 p+q+r=9 , p/(x²-yz)=q/(y²-zx)=r/(z²-xy)
求(px+qy+rz)/(x+y+z) 展开
求(px+qy+rz)/(x+y+z) 展开
1个回答
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设p/(x²-yz)=q/(y²-zx)=r/(z²-xy)=k,则
p=k(x²-yz)
q=k(y²-zx)
r=k(z²-xy)
三式相加得
p+q+r=k(x²+y²+z²-xy-yz-zx)=9
(px+qy+rz)/(x+y+z)
=(xk(x²-yz)+yk(y²-zx)+zk(z²-xy))/(x+y+z)
=k(x³-xyz+y³-xyz+z³-xyz)/(x+y+z)
=k(x³+y³+z³-3xyz)/(x+y+z)
而x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
所以(px+qy+rz)/(x+y+z)=k(x²+y²+z²-xy-yz-zx)=9
p=k(x²-yz)
q=k(y²-zx)
r=k(z²-xy)
三式相加得
p+q+r=k(x²+y²+z²-xy-yz-zx)=9
(px+qy+rz)/(x+y+z)
=(xk(x²-yz)+yk(y²-zx)+zk(z²-xy))/(x+y+z)
=k(x³-xyz+y³-xyz+z³-xyz)/(x+y+z)
=k(x³+y³+z³-3xyz)/(x+y+z)
而x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
所以(px+qy+rz)/(x+y+z)=k(x²+y²+z²-xy-yz-zx)=9
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