2√5+√6=?

直线xcosα+√3y+2=0的倾斜角范围是?A[0,5π/6]
B[π/6,5π/6]
c[π/6,π/2)∪(π/2,5π/6]
D [0,π/6]∪[5π/6,π)帮帮忙谢谢要详细过程,有文字说明,而且π怎么来的?_百度作业帮
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直线xcosα+√3y+2=0的倾斜角范围是?A[0,5π/6]
B[π/6,5π/6]
c[π/6,π/2)∪(π/2,5π/6]
D [0,π/6]∪[5π/6,π)帮帮忙谢谢要详细过程,有文字说明,而且π怎么来的?
直线xcosα+√3y+2=0的倾斜角范围是?A[0,5π/6]
B[π/6,5π/6]
c[π/6,π/2)∪(π/2,5π/6]
D [0,π/6]∪[5π/6,π)帮帮忙谢谢要详细过程,有文字说明,而且π怎么来的?
k=-cosα/√3∈【-√3/3,√3/3】所以倾斜角范围是 [0,π/6]∪[5π/6,π)
Cxcosα+√3y+2=0√3y=-xcosα-2k=-√3cosa/3-1≤cosa≤1√3/3≤k或k≤-√3/3所以k∈[π/6,π/2)∪(π/2,5π/6]
xcosα+√3y+2=0
√3y=-xcosα-2
k=-√3cosa/3
-1≤cosa≤1
-√3/3≤k≤√3/3
也就是说tana∈【-√3/3,√3/3】
a∈[0,π/6]∪[5π/6,π)
是因为k=tana,tana范围在-1到1之间,所以才算出来的√3/3这些值的设M=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),N=1-2+3-4+5-6+…+,求求N/(M+1)²_百度作业帮
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设M=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),N=1-2+3-4+5-6+…+,求求N/(M+1)²
设M=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),N=1-2+3-4+5-6+…+,求求N/(M+1)²
N=1-2+3-4+5-6+…+=-1006M+1=1+1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),利用平方差公司从第二项处理成:1/(1+√2)=(1-√2)/1-2=-(1-√2)、1/(√2+√3)=(√2-√3)/2-3到最后M+1=1-(1-√2)-(√2-√3)-(√3-√4)-…-(√2011-√2012)=√2012N/(M+1)²=1006/(√/2√(6+2√5)分之一等于多少?大根号套小根号怎么算?急耶!_百度作业帮
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√(6+2√5)分之一等于多少?大根号套小根号怎么算?急耶!
√(6+2√5)分之一等于多少?大根号套小根号怎么算?急耶!
6+2√5=5+2√5+1=(√5+1)²所以原式=1/(√5+1)=(√5-1)/(√5+1)(√5-1)=(√5-1)/(5-1)=(1/4)√5-1/4
1/√(6+2√5)=(6-2√5)/(6^2-(2√5)^2)=(6-2√5)/16=(3-√5))/8√32-√8/√2+(2-√3)(2+√3),(2√2/3-√6+√0.5)-(√24+√1/8-√50)√1/2乘√6+√9除√3的值_百度作业帮
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√32-√8/√2+(2-√3)(2+√3),(2√2/3-√6+√0.5)-(√24+√1/8-√50)√1/2乘√6+√9除√3的值
√32-√8/√2+(2-√3)(2+√3),(2√2/3-√6+√0.5)-(√24+√1/8-√50)√1/2乘√6+√9除√3的值
1.原式=4根号2--2+1=--1+2要命号22.原式=(2根号6)/3--根号6+(根号2)/2--2根号6--(根号2)/4+5根号2=--(7根号6)/3+(21根号2)/4.3.原式=根号[(1/2)x6]+根号(9/3)=根号3+根号3=2根号3.
第一题=4√2+4+1=5+4√2
第二题=(2/3 √6-√6+1/2 √2)-(2√6+1/4 √2-5√2)
=-7/3 √6-19/4 √2(5+2√6)^1/2-2^1/2-6^√27+(8^-2/3)^-2_百度作业帮
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(5+2√6)^1/2-2^1/2-6^√27+(8^-2/3)^-2
(5+2√6)^1/2-2^1/2-6^√27+(8^-2/3)^-2
(5+2√6)^1/2-2^1/2-6^√27+(8^-2/3)^-2=√(3+2√6+2)-√2-18√3+(2^-2)^-2=√(3+2)²-√2-18√3+2^4=√3+√2-√2-18√3+16=16-17√3

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