已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2a^2,求x,y,z的取值范围已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2×a^2,求x,y,z的取值范围
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![已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2a^2,求x,y,z的取值范围已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2×a^2,求x,y,z的取值范围](/uploads/image/z/14777442-18-2.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%E5%9D%87%E4%B8%BA%E5%AE%9E%E6%95%B0%2Ca%3E0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3x%2By%2Bz%3Da%2Cx%5E2%2By%5E2%2By%5E2%3D1%2F2a%5E2%2C%E6%B1%82x%2Cy%2Cz%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%E5%9D%87%E4%B8%BA%E5%AE%9E%E6%95%B0%2Ca%3E0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3x%2By%2Bz%3Da%2Cx%5E2%2By%5E2%2By%5E2%3D1%2F2%C3%97a%5E2%2C%E6%B1%82x%2Cy%2Cz%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2a^2,求x,y,z的取值范围已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2×a^2,求x,y,z的取值范围
已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2a^2,求x,y,z的取值范围
已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2×a^2,求x,y,z的取值范围
已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2a^2,求x,y,z的取值范围已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+y^2=1/2×a^2,求x,y,z的取值范围
见4 对称元做主元
x+y+z=a (1)
x^2+y^2+z^2=1/2a^2 (2)
(1)得到:
y+z=a-x (3)
(3)平方得到:y^2+z^2+2yz=x^2-2x+a^2 (4)
(4)-(2):yz=x^2-x+1/4a^2
所以y、z是方程 t^2-(a-x)t+(x^2-x+1/4a^2)=0的两根
所以判别式△=(x-a)...
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x+y+z=a (1)
x^2+y^2+z^2=1/2a^2 (2)
(1)得到:
y+z=a-x (3)
(3)平方得到:y^2+z^2+2yz=x^2-2x+a^2 (4)
(4)-(2):yz=x^2-x+1/4a^2
所以y、z是方程 t^2-(a-x)t+(x^2-x+1/4a^2)=0的两根
所以判别式△=(x-a)^2-4(x^2-x+1/4a^2)>=0
所以0<=x<=2/3a^2
所以x的范围是[0,2/3a^2]
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