实数x、y、z满足x+y+z+8=2√(x-1)+4√(y-2)+6√(z-3),求x+y+z的值是多少?
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实数x、y、z满足x+y+z+8=2√(x-1)+4√(y-2)+6√(z-3),求x+y+z的值是多少?
![实数x、y、z满足x+y+z+8=2√(x-1)+4√(y-2)+6√(z-3),求x+y+z的值是多少?](/uploads/image/z/19311956-44-6.jpg?t=%E5%AE%9E%E6%95%B0x%E3%80%81y%E3%80%81z%E6%BB%A1%E8%B6%B3x%2By%2Bz%2B8%3D2%E2%88%9A%28x-1%29%2B4%E2%88%9A%28y-2%29%2B6%E2%88%9A%28z-3%29%2C%E6%B1%82x%2By%2Bz%E7%9A%84%E5%80%BC%E6%98%AF%E5%A4%9A%E5%B0%91%3F)
x+y+z+8=2√(x-1)+4√(y-2)+6√(z-3)x+y+z+8-2√(x-1)-4√(y-2)-6√(z-3)=0(x-1)-2√(x-1)+1+(y-2)-4√(y-2)+4+(z-3)-6√(z-3)+9=0(√(x-1)-1)^2+(√(y-2)-2)^2+(√(z-3)-3)^2=0(√(x-1)-1)^2>=0,(√(y-2)-2)^2>=0,(√(z-3)-3)^2>=0,而三者的和为0,说明三者均为0√(x-1)-1=0,√(y-2)-2=0,√(z-3)-3=0所以x=2,y=6,z=12x+y+z=2+6+12=20
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