已知实数x,y满足(x-1)^3+211(x-1)=-1和(y-1)^3+211(y-1)=1两式,求x+y的值.
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已知实数x,y满足(x-1)^3+211(x-1)=-1和(y-1)^3+211(y-1)=1两式,求x+y的值.
![已知实数x,y满足(x-1)^3+211(x-1)=-1和(y-1)^3+211(y-1)=1两式,求x+y的值.](/uploads/image/z/620216-8-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9E%E6%95%B0x%2Cy%E6%BB%A1%E8%B6%B3%EF%BC%88x-1%29%5E3%2B211%28x-1%29%3D-1%E5%92%8C%28y-1%29%5E3%2B211%28y-1%29%3D1%E4%B8%A4%E5%BC%8F%2C%E6%B1%82x%2By%E7%9A%84%E5%80%BC.)
将两式相加得
(x-1)^3+(y-1)^3+211(x-1)+211(y-1)=1-1=0
即[(x-1)^3+(y-1)^3]+[211(x-1)+211(y-1)]=0
作因式分解得
(x-1+y-1)[(x-1)^2-(x-1)(y-1)+(y-1)^2]+211(x-1+y-1)=0
即(x+y-2)[(x-1)^2-(x-1)(y-1)+(y-1)^2+211]=0
而f(x)=(x-1)^2-(x-1)(y-1)+(y-1)^2的delta为delta=(y-1)^2-4*(y-1)^2=-3(y-1)^20,f(x)+211>0
因此x+y-2=0
即x+y=2
(x-1)^3+(y-1)^3+211(x-1)+211(y-1)=1-1=0
即[(x-1)^3+(y-1)^3]+[211(x-1)+211(y-1)]=0
作因式分解得
(x-1+y-1)[(x-1)^2-(x-1)(y-1)+(y-1)^2]+211(x-1+y-1)=0
即(x+y-2)[(x-1)^2-(x-1)(y-1)+(y-1)^2+211]=0
而f(x)=(x-1)^2-(x-1)(y-1)+(y-1)^2的delta为delta=(y-1)^2-4*(y-1)^2=-3(y-1)^20,f(x)+211>0
因此x+y-2=0
即x+y=2
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