2003x³=2004y³=2005z³,xyz>0,且³√2003x²+2004y²+2005z³=³√2003+³√2004+³√2005,求1/x+1/y+1/z
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2003x³=2004y³=2005z³,xyz>0,且³√2003x²+2004y²+2005z³=³√2003+³√2004+³√2005,求1/x+1/y+1/z
2003x³=2004y³=2005z³,xyz>0,且³√2003x²+2004y²+2005z³=³√2003+³√2004+³√2005,求1/x+1/y+1/z
2003x³=2004y³=2005z³,xyz>0,且³√2003x²+2004y²+2005z³=³√2003+³√2004+³√2005,求1/x+1/y+1/z
楼下的是不是sb 3的话 x,y,z不都是1
那2003x的平方就不等于2004y的平方就不等于2005z的平方了.这题让我想想再回答你好吗?
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