設√(39+√432)的整數部分為a,小數為b, 求11/a+b+11/a-b+4=?

Sol
1 √(39+√432)
=√(39+2√108)
=√[36+2√(36*3)+3]
=√36+√3
a=7，b=√3－1
2 11/(a+b)+11/(a－b+4)
Sol
a+b=6+√3
a－b+4=7－(√3－1)+4=12－√3
11/(a+b)+11/(a－b+4)
=11/(6+√3)+11/(12－√3)
=11(6－√3)/33+11(12+√3)/(144－3)
=(6－√3)/3+11(12+√3)/141
=(282－47√3)/141+(132+11√3)/141
=(282－47√3)/141+(132+11√3)/141
=(414－36√3)/141
=(138－12√3)/47