這條數好難 中4數學 10分

(a)
P(x+1)(x+2)+Q(x+2)(x +3)+R(x+3)(x+4) = x2 + 8x + 14
代 x = -1:
Q(1)(2) + R(2)(3) = 7
2Q + 6R = 7 ..... (1)
代 x = -2:
R(1)(2) = 2
R = 1
所以 Q = 1/2 根據 (1)
代 x = -3:
P(-2)(-1) = -1
P = -1/2
(b)
-2(x+1)(x+2)+2(x+2)( x+3)+4(x+3)(x+4) = 3x2+20x+29
4[(-1/2)(x+1)(x+2)+(1/2)(x+2)( x+3)+(x+3)(x+4)] = 3x2+20x+29
4(x2 + 8x + 14) = 3x2+20x+29
4x2+32x+56 = 3x2+20x+29
x2+12x+27 = 0
(x+9)(x+3) = 0
x = -9 或 -3

(a)
若P(x+1)(x+2)+Q(x+2)(x+3)+R(x+3)(x+4)恆等x的2次+8x+14 球P,Q,R的值

P(x+1)(x+2)+Q(x+2)(x+3)+R(x+3)(x+4) = x^2+8x+14

When x = -3

P(-2)(-1) = 9 - 24 + 14

P = -1/2

When x = -2

R(1)(2) = 4 - 16 + 14

R = 2/2

R = 1

when x = -1

Q(1)(2) + (1)(2)(3) = 1 - 8 + 14

2Q = 7 - 6

Q = 1/2

So,

-(x + 1)(x + 2) + (x + 2)(x + 3) + 2(x + 3)(x + 4) = x^2 + 8x + 14




(b)由此球-2(x+1)(x+2)+2(x+2)(x+3)=4(x+3)(x+4)=3x的2次+20x+29

-2(x+1)(x+2)+2(x+2)(x+3)+4(x+3)(x+4)=3x^2+20x+29

2[ -(x + 1)(x + 2) + (x + 2)(x + 3) + 2(x + 3)(x + 4)] = 3x^2 + 20x + 29

2x^2 + 16x + 28 = 3x^2 + 20x + 29

x^2 + 4x + 1 = 0

x = [ -4 +/- sprt(12) ]/ 2

x = -2 +/- sprt(3)

a.

P(x+1)(x+2) + Q(x+2)(x+3) + R(x+3)(x+4) ≡ x² + 8x + 14 ..... (*)

由(*), 代x = -2,

P(-2+1)(-2+2) + Q(-2+2)(-2+3) + R(-2+3)(-2+4) = (-2)² + 8(-2) + 14

R(1)(2) = 2

R = 1

由(*), 代x = -3,

P(-3+1)(-3+2) + Q(-3+2)(-3+3) + R(-3+3)(-3+4) = (-3)² + 8(-3) + 14

P(-2)(-1) = -1

P = -1/2

因此, (*)變為 (-1/2)(x+1)(x+2) + Q(x+2)(x+3) + (x+3)(x+4) ≡ x² + 8x + 14 ..... (**)

由(**), 代x = -1,

(-1/2)(-1+1)(-1+2) + Q(-1+2)(-1+3) + (-1+3)(-1+4) = (-1)² + 8(-1) + 14

Q(1)(2) + (2)(3) = 7

Q = 1/2

b.

因此, (**)變為

(-1/2)(x+1)(x+2) + (1/2)(x+2)(x+3) + (x+3)(x+4) ≡ x² + 8x + 14

-2(x+1)(x+2) + 2(x+2)(x+3) + 4(x+3)(x+4) ≡ 4x² + 32x + 56

-2(x+1)(x+2) + 2(x+2)(x+3) + 4(x+3)(x+4) = 3x² + 20x + 29

4x² + 32x + 56 = 3x² + 20x + 29

x² + 12x + 27 = 0

(x + 3)(x + 9) = 0

x = -3 或 x = -9

P(x+1)(x+2)+Q(x+2)(x+3)+R(x+3)(x+4)=x^2+8x+14

put x= -2,-3,-1 and you get the value

2007-03-04 21:55:17 補充：
so P= -1/2,R=1,Q=5

2007-03-04 23:50:46 補充：
計錯數....sorry