初二的二次根式计算题题目 拜托啦
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发布时间：2022-04-24 18:14
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热心网友
时间：2023-10-31 01:36
=3/(2*(2^0.5))+(4/3)*(27/(2^0.5))-3/(6^0.5)
=(3/2+(4/3)*(27)-3/(3^0.5))/(2^0.5)
=(3/2+4*9-(3^0.5))/(2^0.5)
=(75-2*(3^0.5))/(2*2^0.5)
=75/(2*2^0.5)-(3/2)^0.5)
根号0.2m + m分之1根号5m^3 - m根号125分之m
(0.2)^0.5*m+(5m^3)^0.5/m-m*(m/(125^0.5))
=(0.2)^0.5*m+(5m)^0.5-m^2/(5*5^0.5)
=((0.2)^0.5*m*(5*5^0.5)+(5*5^0.5)*(5m)^0.5-m^2)/(5*5^0.5)
=(5m+25*(m)^0.5-m^2)/(5*5^0.5)
6分之1根号5分之8 * （-5根号5分之18）
(1/6)*(8/(5^0.5))*(-5*18/(5^0.5))
=(1/3)*(4/(5^0.5))*(-(5^0.5)*18)
=-(4/3)*18
=-24
根号2x * 根号2y * 根号x
(2x)^0.5*(2y)^0.5*(x^0.5)
=(2x*2y*x)^0.5
=2x*(y^0.5)
（根号mn - 根号n分之m）/根号n分之m (n大于0)
((mn)^0.5-m/(n^0.5))/(m/(n^0.5))
=((mn)^0.5-m/(n^0.5))*(n^0.5)/m
=(n(m^0.5)-m)/(n^0.5)*(n^0.5)/m
=(n(m^0.5)-m)/m
=n/(m^0.5)-1
