F.4 有理數 急！

您好，我是 lop，高興能解答您的問題。

(1)

4/√11-√11/3
= 4(√11)/(√11)(√11) - √11/3
= 4√11/11 - √11/3
= 12√11/33 - 11√11/33
= √11/33

(2a)

化簡

(√3+5)(√3-5)^2
= [(√3)^2-5^2](√3-5)
= (3-5)(√3-5)
= -2(√3-5)
= 10-2√3

(2b) 由此，把1/√3+5的分母有 理化。

1/(√3+5)
= (√3-5)/(√3+5)(√3-5)
= (√3-5)/(-2)
= (5-√3)/2

(3a) 化簡

(√7+√5)(√7-√5)
= (√7)^2-(√5)^2
= 7-5
= 2

(3b) 由此，化簡

√5/(√7+√5)-√7/(√7-√5)
= √5(√7-√5)/(√7+√5)(√7-√5) - √7(√7+√5)/(√7-√5)(√7+√5)
= (√35-5)/2 - (7+√35)/2
= -12/2
= -6

(4a) 化簡

(√7-√2)^2
= √7(√7-√2) - √2(√7-√2)
= 7 - √14 - √14 + 2
= 9 - 2√14

(4b) 由此，化簡

[(√7-√2)+3][(√7-√2)-3]
= (√7-√2)^2 - 3^2
= 9 - 2√14 - 9
= 2√14

2011-09-30 21:13:43 補充：
(2a)

化簡

(√3+5)(√3-5)
= (√3)^2-5^2
= 3 - 25
= -22

1.
4/√11-√11/3
=4/√11*√11/√11-√11/3
=4√11/11-√11/3
=12√11/33-11√11/33
=√11/33

2a
(√3+5)(√3-5)
=√3^2-5^2
=3-25
=-22

2a
1/√3+5
=1/√3+5*(√3-5)/(√3-5)
=(√3-5)/-22
=(5-√3)/22

3a.
(√7+√5)(√7-√5)
=√7^2-√5^2
=7-5
=2

3b
√5/(√7+√5)-√7/(√7-√5)
=[√5(√7-√5)-(√7+√5)√7]/2
=(√35-5-7-√35)/2
=-12/2
=-6

4a
(√7-√2)^2
=7-2√14+2
=9-2√14

4b
[(√7-√2)+3][(√7-√2)-3]
=(√7-√2)^2-9
=9-2√14-9
=-2√14