1.化簡(根號3+根號2)(根號21-根號14)分之根號28=? 2.設f(x)為二次多項式,且滿足f(3)=f(4)=1,f(2)=7,試求f(5)之值 3.方程式x(x+1)(x+2)(x+3)-24=0共有?個實數解?

1.
√28 / [ ( √3 + √2 )( √21 - √14 ) ]
= 2√7 / [ ( √3 + √2 )(√7 )( √3 - √2 ) ]
= 2 / [ ( √3 + √2 )( √3 - √2 ) ]
= 2 / ( 3 - 2 )
= 2 ..... Ans

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2.
因為 f(3) = f(4) = 1 , 故可令:
f(x) = k( x - 3 )( x - 4 ) + 1

f(2) = k( 2 - 3 )( 2 - 4 ) + 1 = k(-1)(-2) + 1 = 2k + 1 = 7
k = 3

f(5) = 3( 5 - 3 )( 5 - 4 ) + 1 = 3*2*1 + 1 = 6 + 1 = 7
Ans: 7

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3.
x(x+1)(x+2)(x+3) - 24 = 0
[ x(x+3) ]*[ (x+1)(x+2) ] - 24 = 0
( x^2 + 3x )( x^2 + 3x + 2 ) - 24 = 0

令 u = x^2 + 3x , 則:
u( u + 2 ) - 24 = 0
u^2 + 2u - 24 = 0
( u + 6 )( u - 4 ) = 0
u = - 6 , 4

當 u = - 6 時:
u = x^2 + 3x = - 6
x^2 + 3x + 6 = 0
判別式 D = 3^2 - 4*1*6 = 9 - 24 < 0
故無實數解

當 u = 4 時:
u = x^2 + 3x = 4
x^2 + 3x - 4 = 0
( x + 4 )( x - 1 ) = 0
x = - 4 , 1
Ans: 有二個實數解 ( x = - 4 , 1 )