设2^a=3^(2b)=6^(2c)=k,则 2=k^(1/a),3=k^(1/2b),6=k^(1/2c),因为2*3=6,所以,k^(1/a)*k(1/2b)=k^(1/2c),即 k^[1/a+1/(2b)]=k^[1/(2c)],所以,1/a+1/(2b)=1/(2c),即 2/a+1/b=1/c.(或者：2bc+ac=ab)