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IFT入學考試題, 求教, 要求詳細教答過程

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• 請問太古股份A, 同太古股份B有咩分別？@1@
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• agnes b 袋真定假@1@

 

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1) if 1/2log y =1+log x,, then y = 100x^2 但點解,同點計算出來 2) 設k 為常數 若y,z為 方程x^2 -3x +k =0 的根 , 則 y^2+ 3z = ? 但點解 同點計算出來 3) 解 9^(x+1/2) - 3^(x+1) = 1/3 + 3^(x-1)

最佳解答:

(1) 1/2log y =1+log x log y = 2 ( 1+log x ) y = 10^( 2 ( 1+log x ) ) y = 10^( 2 + 2log x ) y = ( 10^2 ) ( 10^(2log x) ) y = (100) ( 10^(log(x^2)) ) y = 100 ( 10^log(x^2) ) y = 100 x^2 (2) 設 f(x) = x^2 -3x +k 若y,z為它的根，即f(y) = 0 and f(z) = 0。 f(x) = (x-y)(x-z) f(x) = x^2 - (y+z)x + yz -3 = -(y+z) .... (1) f(y) = y^2 -3y +k = 0 y^2 -3y +k = 0 y^2 + 3z - 3z -3y +k = 0 y^2 + 3z = 3z +3y - k y^2 + 3z = 3(z+y) - k y^2 + 3z = 3(3) - k 由(1) y^2 + 3z = 9 - k (3) 9^(x+1/2) - 3^(x+1) = 1/3 + 3^(x-1) 3^(2x+1) - 3^(x+1) = 3^-1 + 3^(x-1) 3^(2x+1) - 3^(x+1) - 3^-1 - 3^(x-1) = 0 3^(2x+1) - 3^(x+1) - 3^-1 - 3^(x-1) = 0 3(3^x)^2 - 3(3^x) - 1/3 (3^x) - 1/3 = 0 .... (1) Let y = 3^x Sub y into (1), 3y^2 - 3y - 1/3 y - 1/3 = 0 3y^2 - 10/3 y - 1/3 = 0 9 y^2 - 10 y - 1 = 0 By solving the above equation, you have y Sub y into y = 3^x x = log y / log 3

其他解答:

1) 1/2logy=1+logx logy^1/2=log10+logx logy^1/2=log10x y^1/2=10x (y^1/2)^2=(10x)^2 y=100x^2 2) x^2-3x+k=0 y=[3+(9-4k)^1/2]/2 z=[3-(9-4k)^1/2]^2 y^2=(9+6(9-4k)^1/2+9-4k)/4=[18+6(9-4k)^1/2-4k)/4 3z=[9-3(9-4k)^1/2]/2=[18-6(9-4k)^1/2]/4 y^2+3z=(36-4k)/4=9-k 3) 9^(x+1/2)-3^(x+1)=1/3+3^(x-1) 3^(2x+1)-3^(x+1)=3^(-1)+3^(x-1) 3?3^(2x)-3^x?3=3^(-1)+3^x?1/3 設y=3^x 3?y^2-3y=3^(-1)+y/3 9y^2-9y=1+y 9y^2-10y-1=0 b^2-4ac=136 y=[10+/-(136)^1/2]/18 y=[10+/-(2)(17)^1/2]/18 y=[5+-(17)^1/2]/9 3^x=[5+-(17)^1/2]/9