Question

已知x=二分之一(根号7+根号5)y=二分之一(根号7-根号5)求x分之y+y分之x

Answer 1

解:依题设有x=(√7+√5)/2 , y=(√7-√5)/2
所以,原式=y/x + x/y 通分
=(x²+y²)/(xy) 配方
=[(x+y)²-2xy]/(xy) 整体代入
={ [(√7+√5)/2+(√7-√5)/2]²-2×(√7+√5)/2×(√7-√5)/2 }/[(√7+√5)/2×(√7-√5)/2]
=[ (√7)²-(√7²-√5²)/2 ]/[(√7²-√5²)/4]
=[ 7-(7-5)/2 ]/[(7-5)/4]
=(7-1)/(1/2)
=6×2

Answer 2

x=(√7+√5)/2 , y=(√7-√5)/2

原式=y/x + x/y 通分
=(x²+y²)/(xy)
=[(x+y)²-2xy]/(xy)
={ [(√7+√5)/2+(√7-√5)/2]²-2×(√7+√5)/2×(√7-√5)/2 }/[(√7+√5)/2×(√7-√5)/2]
=[ (√7)²-(√7²-√5²)/2 ]/[(√7²-√5²)/4]
=[ 7-(7-5)/2 ]/[(7-5)/4]
=(7-1)/(1/2)
=6×2
=12

Answer 3

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