数学二次根式如何计算
1个回答
展开全部
(1)√3/8-(-3/4√27/2+3√1/6)
=1/4√6+3/8√6-1/2√6
=1/8√6
(2)2/3√9x+6√x/4-2x√1/x
=2√3x+3/2√x-2√x
=5/2√x
(3)2/a√4a+√1/a-2a√1/a^3
=1/a√a+1/a√a-2/a√a
=0
(4)√0.2m+1/m√5m^3-m√125/m
=1/5√5m+√5m-5√5m
=-19/5√5m
(5)√a+b/a-b-√a-b/a+b-√1/a^2-b^2(a>b>0)
=1/(a-b)√(a²-b²)-1/(a+b)√(a²-b²)-1/(a²-b²)√(a²-b²)
=(a+b-a+b-1)/(a²-b²)√(a²-b²)
=(2b+1)/(a²-b²)√(a²-b²)
解不等式
(6)2x+√32<x+√2
2x-x<√2-4√2
x<-3√2
希望能解决您的问题。
=1/4√6+3/8√6-1/2√6
=1/8√6
(2)2/3√9x+6√x/4-2x√1/x
=2√3x+3/2√x-2√x
=5/2√x
(3)2/a√4a+√1/a-2a√1/a^3
=1/a√a+1/a√a-2/a√a
=0
(4)√0.2m+1/m√5m^3-m√125/m
=1/5√5m+√5m-5√5m
=-19/5√5m
(5)√a+b/a-b-√a-b/a+b-√1/a^2-b^2(a>b>0)
=1/(a-b)√(a²-b²)-1/(a+b)√(a²-b²)-1/(a²-b²)√(a²-b²)
=(a+b-a+b-1)/(a²-b²)√(a²-b²)
=(2b+1)/(a²-b²)√(a²-b²)
解不等式
(6)2x+√32<x+√2
2x-x<√2-4√2
x<-3√2
希望能解决您的问题。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询