推 gagaRicky :太厲害了!!高手~謝謝 01/10 10:59
※ 引述《gagaRicky (Ricky)》之銘言:
: 1. x^2+ 2log5 + log5/2 =0兩根為a,b
: 求10^a+10^b為多少
log(5/2)=log5-log2 = log5-(1-log5)=2log5-1
故x^2+2log5+(2log5-1)= (x+2log5-1)(x+1)=0
得兩根為1-2log5, -1
因此10^a+10^b = 10^(1-2log5) + 10^(-1) =10^(log(2/5)) + 1/10
= 2/5 + 1/10 = 1/2
: 2. y=logx 與 斜率為1/2的直線交於AB兩點
: 且x座標為a,b
: __
: 求a,b的關係式 又AB=2根號5 a,b 為何?
設直線為y=(1/2)x + k
則(1/2)x + k =logx 之兩根為a,b
故(1/2)a+ k = loga
(1/2)b+ k = logb
兩式相減得(1/2)(a-b)=loga-logb=log(a/b)....(1)
設A(a,loga), B(b,logb), a>b
__2
則AB = (a-b)^2+(loga-logb)^2 = (a-b)^2+[(1/2)(a-b)]^2
= (5/4)(a-b)^2 = 20
得(a-b)^2= 16 => a-b=4
代回(1)得2=log(a/b) => a/b=100 => a=100b
因此解得a=400/99, b=4/99
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