Re: [中學] 嘉中科學班考題..
※ 引述《paggei (XD)》之銘言:
: 2. 設(a + sqrt(a^2 - 9))(b + sqrt(b^2 + 5)) = 12
: 求a*sqrt(b^2 + 5) + b*sqrt(a^2 - 9)的值？
: Ans:33/8
方便起見，令sqrt(a^2 - 9)=X ，sqrt(b^2 + 5)=Y
=> (a+X)(b+Y)=12 ， 求 aY+bX=?
=> (a+X)(a-X)(b+Y)(Y-b)=12(a-X)(Y-b)
=> 9*5 =12(aY-ab-XY+bX)
=> 15/4 = 12-2*(ab+XY) (ab+aY+bX+XY=12)
=> ab+XY = 33/8
=> aY+bX = 63/8
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