maths(F.4)

( 1 ) ( log 8 + log 4 ) / 1 / 2 log 2

= 2 log ( 8 )( 4 ) / log 2

= 2 log 32 / log 2

= 2 log 25 / log 2

= 10 log 2 / log 2

= 10

( 2a ) x2 + x - k2 – k

= ( x + k )( x – k ) + ( x – k )

= ( x – k )( x + k + 1 )

b) ( x + 4)( x – 3 ) = ( k + 4 )( k – 3 )

x2 – 3x + 4x – 12 = k2 – 3k + 4k – 12

x2 + x = k2 + k

x2 + x – k2 – k = 0

所以,

( x – k )( x + k + 1 ) = 0

x = k 或 – k – 1

( 3 ) 代x = 4,

( - 4 )2 - ( 6 + k )( - 4 ) + k = 0

16 + 24 + 4k + k = 0

5k = -40

k = -8

於是,

x2 - (6 -8 ) x - 8 = 0

x2 + 2x – 8 = 0

( x + 4 )( x – 2 ) = 0

x = -4 或 2

上述方程另一個根是2。

( 4a ) 設f ( x ) = x3 – x2 - 3x – 1

f ( -1 ) = ( -1 )3 – ( -1 )2 – 3 ( - 1 ) – 1

= 0

所以x + 1是x3 – x2 - 3x - 1的因式

( b ) x3 – x2 - 3x – 1 = 0

( x + 1 )( x2 – 2x – 1 ) = 0

x = - 1 或 1 ±√2

( 5a ) y = mx + 2 --- ( 1 )

y = x2 – x + 6 --- ( 2 )

代( 1 )入( 2 ),

mx + 2 = x2 – x + 6

x2 – ( m + 1 ) x + 4 = 0

假如兩者相切, m + 1 = 5 或 – 5

所以, m = 4 或 – 6

b) 若取m為負值,

x2 – ( -6 + 1 ) x + 4 = 0

x2 + 5x + 4 = 0

( x + 4 )( x + 1 ) = 0

x = -4 或 –1

y = -6 ( - 4 ) + 2或 – 6 ( - 1 ) + 2

= 26或 8

所以切點坐標: ( - 4, 26 ) , ( - 1 , 8 )

( 6 ) y = x2 + 3x – 5 --- ( 1 )

x + y = k

y = k – x --- ( 2 )

代( 1 )入( 2 ),

x2 + 3x – 5 = k – x

x2 + 4x + ( - 5 – k ) = 0

由於只有一對解, △ = 0

42 – 4 ( 1 )( - 5 – k ) = 0

1 + 5 + k = 0

k = -6

log8+log4 / 1/2 log2的值
ans:log8+log4/1/2log2
=log(8x4)/1/2log2
=2log32/log2
=2log2^5/log2
=10log2/log2
=10


若-4是方程x^2-(6+k)x+k=0的其中一個根,求
k的值
上述方程另一個根

ans:將-4代入方程

(-4)^2-(6+k)(-4)+k=o
16+24+4k+k=0
5k=-40
k=8

x^2-(6+8)x+8=0
x^2-48x+8=0
(x-4)(x+12)=0

x=4 or x=-12


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