周末截止1.√（√48-√45）2.√（21-4√5+8√3-4√15）3.（√5+2√7+3）/(√35+3√5+3√7+7)4.若a=1995^2+1995^2*1996^2+1996^2,求√a5.√（1+2005^2+2005^2/2006^2)-1/2006

Question

周末截止
1.√（√48-√45）
2.√（21-4√5+8√3-4√15）
3.（√5+2√7+3）/(√35+3√5+3√7+7)
4.若a=1995^2+1995^2*1996^2+1996^2,求√a
5.√（1+2005^2+2005^2/2006^2)-1/2006

Answer 1

1,
[(48)^(1/2) - (45)^(1/2)]^(1/2) = [4*3^(1/2) - 3*5^(1/2)]^(1/2)
= 3^(1/4)[4 - (15)^(1/2)]^(1/2) = [3/4]^(1/4)[8 - 2(15)^(1/2)]^(1/2)
= [3/4]^(1/4){[5^(1/2) - 3^(1/2)]^2}^(1/2) = [3/4]^(1/4)[5^(1/2) - 3^(1/2)]
= 3^(1/4)[(5/2)^(1/2) - (3/2)^(1/2)]
2,
[21 - 4*5^(1/2) + 8*3^(1/2) - 4*(15)^(1/2)]^(1/2)
={21 + 2[2*(12)^(1/2) - 2*5^(1/2) - (12)^(1/2)*5^(1/2)]}^(1/2)
={12+5+2^2 + 2[2*(12)^(1/2) - 2*5^(1/2) - (12)^(1/2)*5^(1/2)]}^(1/2)
= {[(12)^(1/2)+2-5^(1/2)]^2}^(1/2)
= (12)^(1/2)+2-5^(1/2)
= 2*3^(1/2) + 2 - 5^(1/2)
3,
[5^(1/2) + 2*7^(1/2) + 3]/[(35)^(1/2) + 3*5^(1/2) + 3*7^(1/2) + 7]
= [5^(1/2) + 2*7^(1/2) + 3]/{[7^(1/2) + 3]5^(1/2) + 7^(1/2)[7^(1/2) + 3]}
= [5^(1/2)+7^(1/2) + 7^(1/2)+3]/{[7^(1/2) + 3][5^(1/2) + 7^(1/2)]}
= 1/[7^(1/2)+3] + 1/[5^(1/2)+7^(1/2)]
= [3-7^(1/2)]/2 + [7^(1/2)-5^(1/2)]/2
= [3 - 5^(1/2)]/2
4,
a^(1/2) = [1995^2 + 1995^2*1996^2 + 1996^2]^(1/2)
= [1995^2*1996^2-2*1995^2*1996+1995^2] + 2*1996*1995^2+1996^2]^(1/2)
= [1995^4 + 2*1995^2*1996 + 1996^2]^(1/2)
= {[1995^2 + 1996]^2}^(1/2)
= 1995^2 + 1996
5,
{1 + 2005^2 + 2005^2/2006^2}^(1/2) - 1/2006
= {1-2*2005/2006+2005^2/2006^2 +2*2005/2006 + 2005^2}^(1/2) - 1/2006
= {(1-2005/2006)^2 + 2*2005/2006 + 2005^2}^(1/2) - 1/2006
= {1/2006^2 + 2*2005/2006 + 2005^2}^(1/2) - 1/2006
= {[1/2006 + 2005]^2}^(1/2) - 1/2006
= 1/2006 + 2005 - 1/2006
= 2005