98配方數學

1.
因x^2-2x+a=2之兩根相等所已必可配成完全平方式
x^2-2x+a-2=0=>x^2-2x+1=0=>(x-1)^2=0
a-2=1 => a=3 x=1
ax^2-2x-56=0 => 3x^2-2x-56=0 =>(3x-14)(x+4)=0 =>x=14/3or-4

2.
(1)
9x+3a-2=6x+1 => 9x-6x+3a-2-1=0 =>3x+3a-3=0 =>x+a-1=0 =>x=1-a
(2)
(3x+2)(x-4)=(x-4)^2 =>3x^2-10x-8=x^2-8x+16 =>2x^2-2x-24=0
=>x^2-x-12=0 =>(x+3)(x-4)=0 =>x=-3or4
(3)
x=1-a
令x=-3 -3=1-a => a=4
令x=4 4=1-a =>a=-3

3.
x=3代入x^2-2ax+3a=0 =>3^2-6a+3a=0 =>-3a=-9 =>a=3
a代入x^2-3ax+2a=0 =>x^2-6x+6=0 (方程式) =>(x-3)^2-3=0 =>x=3+-√3

4.
1-√2代入(x-1)^2=2a =>2=2a =>a=1
(x-1)^2-2=0 =>x=1+√2

5.
√3/(√3+√2)=√3(√3-√2)/(√3+√2)(√3-√2)=3-√6

6.
x=√3/(√3+√2)
[√3/(√3+√2)]^2-6(√3/(√3+√2))+3=(3-√6)^2-6(3-√6)+3=0

7.
(2x^3-12x^2+6x+7)/(x^2-6x+3)=2x......7(餘式)
f(x)=f[√3/(√3+√2)]=f(3-√6)=2(3-√6)^2-12(3-√6)^2+6(3-√6)+7
=2(3-√6)[(3-√6)-6(3-√6)+3]+7
=2(3-√6)[(3-√6)(1-6)+3]+7
=2(3-√6)[(3-√6)(-5)+3]+7
=2(3-√6)[-12+5√6]+7
=2(-12+5√6)(3-√6)+7
=2(-36+12√6+15√6-30)+7
=2(-66+27√6)
=-132+54√6


有問題再補充

2009-12-07 23:07:11 補充：
補充
2.
(3)
a的範圍
-3<1-a<4 => -4<-a<3 =>4>a>-3

2009-12-09 20:11:45 補充：
3.
更正:
x^2-9x+6=0 (方程式) =>(x-3)^2-3=0 =>x=3+-√3
為-9x

7.
更正:
f(x)=f[√3/(√3+√2)]=f[3-√6]=2(3-√6)^3-12(3-√6)^2+6(3-√6)+7
=2(3-√6)[(3-√6)^2-6(3-√6)+3]+7
=2(3-√6)[9-6√6+6-18+6√6+3]+7
=2(3-√6)[-6]+7
=(6-2√6)[-6]+7
=-36+12√6+7
=-29+12√6

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算錯2題, 刪了!~~