✔ 最佳答案
( 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
