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数学
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已知三次立方根号x等于4,根号√2y-z+1+(z-3)²=0,求√x-y³-z³的值
人气:300 ℃ 时间:2019-11-06 10:04:41
解答
³√x=4
x=64
√(2y-z+1)+(z-3)²=0
√(2y-z+1)=0,(z-3)²=0
2y-z+1=0,z-3=0
2y-z+1=0,z=3
2y-3+1=0
2y-2=0
y-1=0
y=1
所以x=64,y=1,z=3
√(x-y³-z³)
=√(64-1³-3³)
=√(64-1-27)
=√36
=6有点复杂。。。不过谢谢~
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