(a) − 1 − 5 5 + 1 = − ( 1 + 5 ) 5 + 1 = − 1 \dfrac{-1-\sqrt5}{\sqrt5+1}=\dfrac{-(1+\sqrt5)}{\sqrt5+1}=-1 5 + 1 − 1 − 5 = 5 + 1 − ( 1 + 5 ) = − 1 1 − 5 5 − 1 = − ( 5 − 1 ) 5 − 1 = − 1 \dfrac{1-\sqrt5}{\sqrt5-1}=\dfrac{-(\sqrt5-1)}{\sqrt5-1}=-1 5 − 1 1 − 5 = 5 − 1 − ( 5 − 1 ) = − 1 1 1 1 ( 5 − 1 ) 2 ( 1 + 5 ) 2 = 6 − 2 5 6 + 2 5 = 3 − 5 3 + 5 = ( 3 − 5 ) 2 9 − 5 = 7 − 3 5 2 \dfrac{(\sqrt5-1)^2}{(1+\sqrt5)^2}=\dfrac{6-2\sqrt5}{6+2\sqrt5}=\dfrac{3-\sqrt5}{3+\sqrt5}=\dfrac{(3-\sqrt5)^2}{9-5}=\dfrac{7-3\sqrt5}{2} ( 1 + 5 ) 2 ( 5 − 1 ) 2 = 6 + 2 5 6 − 2 5 = 3 + 5 3 − 5 = 9 − 5 ( 3 − 5 ) 2 = 2 7 − 3 5 1 − 7 − 3 5 2 = − 5 + 3 5 2 1-\dfrac{7-3\sqrt5}{2}=\dfrac{-5+3\sqrt5}{2} 1 − 2 7 − 3 5 = 2 − 5 + 3 5 1 5 \tfrac1{\sqrt5} 5 1 5 \sqrt5 5 − 5 + 3 5 2 ⋅ 5 = 15 − 5 5 2 . \frac{-5+3\sqrt5}{2}\cdot\sqrt5=\frac{15-5\sqrt5}{2}. 2 − 5 + 3 5 ⋅ 5 = 2 15 − 5 5 . − 1 − 15 − 5 5 2 = 5 5 − 17 2 -1-\dfrac{15-5\sqrt5}{2}=\dfrac{5\sqrt5-17}{2} − 1 − 2 15 − 5 5 = 2 5 5 − 17
(b) 3 ( 3 − 1 ) 2 = 3 4 − 2 3 = 3 ( 4 + 2 3 ) 16 − 12 = 6 + 3 3 2 \dfrac{3}{(\sqrt3-1)^2}=\dfrac{3}{4-2\sqrt3}=\dfrac{3(4+2\sqrt3)}{16-12}=\dfrac{6+3\sqrt3}{2} ( 3 − 1 ) 2 3 = 4 − 2 3 3 = 16 − 12 3 ( 4 + 2 3 ) = 2 6 + 3 3 2 + 6 6 − 2 = 2 ( 1 + 3 ) 2 ( 3 − 1 ) = ( 1 + 3 ) 2 2 = 2 + 3 \dfrac{\sqrt2+\sqrt6}{\sqrt6-\sqrt2}=\dfrac{\sqrt2(1+\sqrt3)}{\sqrt2(\sqrt3-1)}=\dfrac{(1+\sqrt3)^2}{2}=2+\sqrt3 6 − 2 2 + 6 = 2 ( 3 − 1 ) 2 ( 1 + 3 ) = 2 ( 1 + 3 ) 2 = 2 + 3 ( 2 + 6 ) 2 ( 3 + 1 ) 2 = 8 + 4 3 4 + 2 3 = 4 ( 2 + 3 ) 2 ( 2 + 3 ) = 2 \dfrac{(\sqrt2+\sqrt6)^2}{(\sqrt3+1)^2}=\dfrac{8+4\sqrt3}{4+2\sqrt3}=\dfrac{4(2+\sqrt3)}{2(2+\sqrt3)}=2 ( 3 + 1 ) 2 ( 2 + 6 ) 2 = 4 + 2 3 8 + 4 3 = 2 ( 2 + 3 ) 4 ( 2 + 3 ) = 2 6 + 3 3 2 + ( 2 + 3 ) − 2 = 3 + 3 3 2 + 3 = 3 + 5 3 2 \dfrac{6+3\sqrt3}{2}+(2+\sqrt3)-2=3+\dfrac{3\sqrt3}{2}+\sqrt3=3+\dfrac{5\sqrt3}{2} 2 6 + 3 3 + ( 2 + 3 ) − 2 = 3 + 2 3 3 + 3 = 3 + 2 5 3
(c) ( 5 + 3 ) 2 ( 5 − 3 ) 2 = ( ( 5 + 3 ) 2 5 − 3 ) 2 = ( 4 + 15 ) 2 = 31 + 8 15 \dfrac{(\sqrt5+\sqrt3)^2}{(\sqrt5-\sqrt3)^2}=\left(\dfrac{(\sqrt5+\sqrt3)^2}{5-3}\right)^2=(4+\sqrt{15})^2=31+8\sqrt{15} ( 5 − 3 ) 2 ( 5 + 3 ) 2 = ( 5 − 3 ( 5 + 3 ) 2 ) 2 = ( 4 + 15 ) 2 = 31 + 8 15 ( 5 − 3 ) 2 ( 5 + 3 ) 2 = ( 4 − 15 ) 2 = 31 − 8 15 \dfrac{(\sqrt5-\sqrt3)^2}{(\sqrt5+\sqrt3)^2}=(4-\sqrt{15})^2=31-8\sqrt{15} ( 5 + 3 ) 2 ( 5 − 3 ) 2 = ( 4 − 15 ) 2 = 31 − 8 15 16 15 16\sqrt{15} 16 15 2 − 2 = 2 ( 2 − 1 ) 2-\sqrt2=\sqrt2(\sqrt2-1) 2 − 2 = 2 ( 2 − 1 ) 3 + 3 = 3 ( 3 + 1 ) 3+\sqrt3=\sqrt3(\sqrt3+1) 3 + 3 = 3 ( 3 + 1 ) ( 2 − 2 ) 2 ( 3 + 3 ) 2 = 6 ( 2 − 1 ) 2 ( 3 + 1 ) 2 (2-\sqrt2)^2(3+\sqrt3)^2=6\,(\sqrt2-1)^2(\sqrt3+1)^2 ( 2 − 2 ) 2 ( 3 + 3 ) 2 = 6 ( 2 − 1 ) 2 ( 3 + 1 ) 2 ( 2 − 1 ) 2 ( 3 + 1 ) 2 ( 2 − 2 ) 2 ( 3 + 3 ) 2 = 1 6 . \frac{(\sqrt2-1)^2(\sqrt3+1)^2}{(2-\sqrt2)^2(3+\sqrt3)^2}=\frac16. ( 2 − 2 ) 2 ( 3 + 3 ) 2 ( 2 − 1 ) 2 ( 3 + 1 ) 2 = 6 1 . 16 15 − 1 6 16\sqrt{15}-\dfrac16 16 15 − 6 1
(d) 3 − 5 3 + 5 = ( 3 − 5 ) 2 3 − 5 = 15 − 4 \dfrac{\sqrt3-\sqrt5}{\sqrt3+\sqrt5}=\dfrac{(\sqrt3-\sqrt5)^2}{3-5}=\sqrt{15}-4 3 + 5 3 − 5 = 3 − 5 ( 3 − 5 ) 2 = 15 − 4 1 − ( 15 − 4 ) = 5 − 15 1-(\sqrt{15}-4)=5-\sqrt{15} 1 − ( 15 − 4 ) = 5 − 15 3 − 3 = 3 ( 3 − 1 ) 3-\sqrt3=\sqrt3(\sqrt3-1) 3 − 3 = 3 ( 3 − 1 ) ( 3 − 3 ) ( 5 − 15 ) 3 − 1 = 3 ( 5 − 15 ) = 5 3 − 3 5 . \frac{(3-\sqrt3)(5-\sqrt{15})}{\sqrt3-1}=\sqrt3\,(5-\sqrt{15})=5\sqrt3-3\sqrt5. 3 − 1 ( 3 − 3 ) ( 5 − 15 ) = 3 ( 5 − 15 ) = 5 3 − 3 5 . 1 + ( 5 3 − 3 5 ) = 1 + 5 3 − 3 5 1+(5\sqrt3-3\sqrt5)=1+5\sqrt3-3\sqrt5 1 + ( 5 3 − 3 5 ) = 1 + 5 3 − 3 5 ( 3 + 1 ) 2 = 4 + 2 3 (\sqrt3+1)^2=4+2\sqrt3 ( 3 + 1 ) 2 = 4 + 2 3 34 + 22 3 − 12 5 − 6 15 34+22\sqrt3-12\sqrt5-6\sqrt{15} 34 + 22 3 − 12 5 − 6 15 ( 3 + 3 ) 2 = 12 + 6 3 (3+\sqrt3)^2=12+6\sqrt3 ( 3 + 3 ) 2 = 12 + 6 3 46 + 28 3 − 12 5 − 6 15 . 46+28\sqrt3-12\sqrt5-6\sqrt{15}. 46 + 28 3 − 12 5 − 6 15 .
( a ) 5 5 − 17 2 ( b ) 3 + 5 3 2 ( c ) 16 15 − 1 6 ( d ) 46 + 28 3 − 12 5 − 6 15 \boxed{(a)\ \dfrac{5\sqrt5-17}{2}\qquad (b)\ 3+\dfrac{5\sqrt3}{2}\qquad (c)\ 16\sqrt{15}-\dfrac16\qquad (d)\ 46+28\sqrt3-12\sqrt5-6\sqrt{15}} ( a ) 2 5 5 − 17 ( b ) 3 + 2 5 3 ( c ) 16 15 − 6 1 ( d ) 46 + 28 3 − 12 5 − 6 15