恆等式求值4題

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(1)

3x²+ 7x + 2 ≡ a(x+1)² + b(x+1) + c

3x²+ 7x + 2
= ( 3x²+ 3x ) + 4x + 2
= ( 3x²+ 3x ) + ( 4x + 4 ) - 2
= 3x(x+1) + 4(x+1) - 2
= (3x+4)(x+1) - 2
= [(3x+3)+1](x+1) - 2
= 3(x+1)(x+1) + 1(x+1) - 2
= 3(x+1)²+ 1(x+1) - 2

∴ a = 3 , b = 1 , c = -2

(2)

3x²+ 7x + 2 ≡ a(x+2)² + b(x+2) + c

3x²+ 7x + 2
= ( 3x²+ 6x ) + x + 2
= ( 3x²+ 6x ) + ( x + 2 )
= 3x(x+2) + 1(x+2)
= (3x+1)(x+2)
= [(3x+6)-5](x+2)
= 3(x+2)(x+2) - 5(x+2)
= 3(x+2)²- 5(x+2) + 0

∴ a = 3 , b = -5 , c = 0

(3)

2(x+1)²+ 6(x+1) + 1 ≡ a(x-2)²+ b(x-2) + c

2(x+1)²+ 6(x+1) + 1
= 2(x²+2x+1) + 6x + 6 + 1
= 2x²+ 4x + 2 + 6x + 7
= 2x²+ 10x + 9
= ( 2x²- 4x ) + 14x + 9
= ( 2x²- 4x ) + ( 14x - 28 ) + 37
= 2x(x-2) + 14(x-2) + 37
= (2x+14)(x-2) + 37
= [(2x-4)+18](x-2) + 37
= 2(x-2)(x-2) + 18(x-2) + 37
= 2(x-2)²+ 18(x-2) + 37

∴ a = 2 , b = 18 , c = 37

(4)

3(x-1)³+ 5(x-1)²+ 2(x-1) + 9 ≡ a(x+1)³+ b(x+1)²+ c(x+1) + d

3(x-1)³+ 5(x-1)²+ 2(x-1) + 9
= 3(x³-3x²+3x-1) + 5(x²-2x+1) + 2x - 2 + 9
= 3x³- 9x²+ 9x - 3 + 5x²- 10x + 5 + 2x + 7
= 3x³- 4x²+ x + 9
= ( 3x³+ 3x²) - 7x²+ x + 9
= ( 3x³+ 3x²) - ( 7x²+ 7x ) - 6x + 9
= ( 3x³+ 3x²) - ( 7x²+ 7x ) - ( 6x + 6 ) + 3
= 3x²(x+1) - 7x(x+1) - 6(x+1) +3
= (3x²-7x-6)(x+1) + 19
= [(3x²+3x)-10x-6](x+1) + 19
= [(3x²+3x)-(10x+10)+4](x+1) + 19
= [3x(x+1)-10(x+1)+4](x+1) + 19
= (3x-1)(x+1)(x+1) + 4(x+1) + 19
= [(3x+3)-4](x+1)²+ 4(x+1) + 19
= 3(x+1)(x+1)²- 4(x+1)²+ 4(x+1) + 19
= 3(x+1)³- 4(x+1)²+ 4(x+1) + 19

∴ a = 3 , b = -4 , c = 4 , d = 19

2011-11-16 19:19:39 補充：
corr. (4)

(4)

3(x-1)³+ 5(x-1)²+ 2(x-1) + 9 ≡ a(x+1)³+ b(x+1)²+ c(x+1) + d

3(x-1)³+ 5(x-1)²+ 2(x-1) + 9
= 3(x³-3x²+3x-1) + 5(x²-2x+1) + 2x - 2 + 9
= 3x³- 9x²+ 9x - 3 + 5x²- 10x + 5 + 2x + 7
= 3x³- 4x²+ x + 9

2011-11-16 19:19:45 補充：
= ( 3x³+ 3x²) - 7x²+ x + 9
= ( 3x³+ 3x²) - ( 7x²+ 7x ) + 8x + 9
= ( 3x³+ 3x²) - ( 7x²+ 7x ) + ( 8x + 8 ) + 1
= 3x²(x+1) - 7x(x+1) + 8(x+1) + 1
= (3x²-7x+8)(x+1) + 1
= [(3x²+3x)-10x+8](x+1) + 1
= [(3x²+3x)-(10x+10)+18](x+1) + 1

2011-11-16 19:23:13 補充：
= [(3x²+3x)-(10x+10)+18](x+1) + 1
= [3x(x+1)-10(x+1)+18](x+1) + 1
= (3x-10)(x+1)(x+1) + 18(x+1) + 1
= [(3x+3)-13](x+1)²+ 18(x+1) + 1
= 3(x+1)(x+1)²- 13(x+1)²+ 18(x+1) + 1
= 3(x+1)³- 13(x+1)²+ 18(x+1) + 1

∴ a = 3 , b = -13 , c = 18 , d = 1

1.
3x² + 7x + 2 ≡ a(x+1)² + b(x+1) + c
from R.H.S
a(x+1)² + b(x+1) + c
= ax² + 2ax + a + bx + b + c
= ax² + (2a+b)x + (a+b+c)
as it is a identity, just simply comparing the coefficient of both sides
thus, a=3, 2a+b=7, a+b+c=2
=> a=3, b=1, c=-2

2.
3x² + 7x + 2 ≡ a(x+2)² + b(x+2) + c
from R.H.S
a(x+2)² + b(x+2) + c
= ax² + 4ax + 4a + bx + 2b + c
= ax² + (4a+b)x + (4a+2b+c)
as it is a identity, just simply comparing the coefficient of both sides
thus, a=3, 4a+b=7, 4a+2b+c=2
=> a=3, b=-5, c=0

3.
2(x+1)² + 6(x+1) + 1 ≡ a(x-2)² + b(x-2) + c
from L.H.S
2(x+1)² + 6(x+2) + 1
= 2x² + 4x + 2 + 6x + 12 + 1
= 2x² + 10x + 15
from R.H.S
a(x-2)² + b(x-2) + c
= ax² - 4ax + 4a + bx - 2b + c
= ax² + (-4a+b)x + (4a-2b+c)
as it is a identity, just simply comparing the coefficient of both sides
thus, a=2, -4a+b=10, 4a-2b+c=15
=> a=2, b=18, c=43

4.
3(x-1)³ + 5(x-1)² + 2(x-1) + 9 ≡ a(x+1)³ + b(x+1)² + c(x+1) + d
from L.H.S
3(x-1)³ + 5(x-1)² + 2(x-1) + 9
= 3x³-9x²+9x-3 + 5x²-10x+5 + 2x-2 + 9
= 3x³ - 4x² + x + 9
from R.H.S
a(x+1)³ + b(x+1)² + c(x+1) + d
= ax³+3ax²+3ax+a + bx²+2bx+b + cx+c + d
= ax³ + (3a+b)x² + (3a+2b+c)x + (a+b+c+d)
as it is a identity, just simply comparing the coefficient of both sides
thus, a=3 , 3a+b=-4, 3a+2b+c=1, a+b+c+d=9
=> a=3, b=-13, c=18, d=1




2011-11-17 13:35:25 補充：
corr 3.
2(x+1)² + 6(x+1) + 1 ≡ a(x-2)² + b(x-2) + c
from L.H.S
2(x+1)² + 6(x+1) + 1
= 2x² + 4x + 2 + 6x + 6 + 1
= 2x² + 10x + 9
from R.H.S
= ax² + (-4a+b)x + (4a-2b+c)
as it is a identity, just simply comparing the coefficient of both sides
thus, a=2, -4a+b=10, 4a-2b+c=9
=> a=2, b=18, c=37

My solution:
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-167.jpg