已知x=y+z+2,x²=y²+z²+2yz+200试求(1)x+y+z的值(2)x的值
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![已知x=y+z+2,x²=y²+z²+2yz+200试求(1)x+y+z的值(2)x的值](/uploads/image/z/8566390-46-0.jpg?t=%E5%B7%B2%E7%9F%A5x%3Dy%2Bz%2B2%2Cx%26%23178%3B%3Dy%26%23178%3B%2Bz%26%23178%3B%2B2yz%2B200%E8%AF%95%E6%B1%82%EF%BC%881%EF%BC%89x%2By%2Bz%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89x%E7%9A%84%E5%80%BC)
已知x=y+z+2,x²=y²+z²+2yz+200试求(1)x+y+z的值(2)x的值
已知x=y+z+2,x²=y²+z²+2yz+200试求(1)x+y+z的值(2)x的值
已知x=y+z+2,x²=y²+z²+2yz+200试求(1)x+y+z的值(2)x的值
x=y+z+2,
平方得到
x^2=x^2+y^2+4+2yz+4y+4z
然后减去x²=y²+z²+2yz+200得到
4y+4z=196
所以y+z=49
所以x=y+z+2=51
所以x+y+z=100
x=y+z+2=51
100 x=51
x=y+z+2,x²=y²+z²+2yz+200
x²=(y+z+2)²=y²+z²+2yz+4y+4z+4=y²+z²+2yz+200
4y+4z+4=200
y+z=49
x=y+z+2=51
x+y+z=100