Answer:
To solve this problem we should multiply both numerator and denominator by ( 5 + 1 ) ( 5 + 1 ) \frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}+1)} ( 5 + 1 ) ( 5 + 1 )
( 5 + 1 ) ( 5 − 1 ) ∗ ( 5 + 1 ) ( 5 + 1 ) = ( 5 ∗ 5 + 5 + 5 + 1 ) ( 5 ∗ 5 + 5 − 5 − 1 ) = ( 5 + 2 5 + 1 ) ( 5 − 1 ) \frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}-1)}*\frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}+1)}=\frac{(\sqrt{\smash[b]{5}}*\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}*\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}-\sqrt{\smash[b]{5}}-1)}=\frac{(5+2\sqrt{\smash[b]{5}}+1)}{(5-1)} ( 5 − 1 ) ( 5 + 1 ) ∗ ( 5 + 1 ) ( 5 + 1 ) = ( 5 ∗ 5 + 5 − 5 − 1 ) ( 5 ∗ 5 + 5 + 5 + 1 ) = ( 5 − 1 ) ( 5 + 2 5 + 1 )
⟹ ( 5 + 2 5 + 1 ) ( 5 − 1 ) = ( 6 + 2 5 ) ( 4 ) = 2 ( 3 + 5 ) 4 = 3 + 5 2 \implies \frac{(5+2\sqrt{\smash[b]{5}}+1)}{(5-1)}=\frac{(6+2\sqrt{\smash[b]{5}})}{(4)}=\frac{2(3+\sqrt{\smash[b]{5}})}{4}=\frac{3+\sqrt{\smash[b]{5}}}{2} ⟹ ( 5 − 1 ) ( 5 + 2 5 + 1 ) = ( 4 ) ( 6 + 2 5 ) = 4 2 ( 3 + 5 ) = 2 3 + 5
Answer:
3 + 5 2 \frac{3+\sqrt{\smash[b]{5}}}{2} 2 3 + 5
