F.2 MATHS 3題 (快...)

1)Ax(x-1)(x+2)+Bx(x-1)+Cx+D = 5x* + 3X^ - 10x + 6

Ax(x-1)(x+2)+Bx(x-1)+Cx+D

=Ax(x^+x-2)+Bx^-Bx+Cx+D

=Ax*+Ax^-2Ax+Bx^+(C-B)x+D

=Ax*+(A+B)x^+(C-B-2A)x+D

SO,D=6

A=5

A+B=3

5+B=3

B=-2

C-B-2A=-10

C-(-2)-2(5)=-10

C=-10+10-4

C=-4

so,A=5,B=-2,C=-4,D=6


2)54(46)

=(50+4)(50-4)

=50^-4^

=2500-16

=2484


3)102^ - 16(108) + 63

=102^ - 16(102+6) + 63

=102^ - 16(102) + 63 - 96

=(102-9)(102-7)-96

=(93)(95)-96

=(94-1)(94+1)-96

=94^-1-96

=(100-6)^-97

=100^-2(100)(6)-36-97

=10000-1200-130

=8670

1)Ax(x-1)(x+2)+Bx(x-1)+Cx+D = 5x³ + 3X² - 10x + 6

Ax³-Ax²+Bx²-Bx+Cx+D=5x³ + 3X² - 10x + 6

A=5 B=8 C=-2 D=6
-------------------------------------------------------------------------------------------------------
2)54(46)

=(50+4)(50-4)

=2500+200-200-16

=2516
-------------------------------------------------------------------------------------------------------
3)102² - 16(108) + 63

=102² - (62-46)(62+46) + 63

=102² - (3844+2852-2852-2116) + 63

=102² - 1728 + 63

=(100+2)² - 1728 + 63

=(10000+200-200+4) - 1728 + 63

=10004 - 1728 + 63

=8339

1) A=5 B=-2 C=-10 D=6
2) 2484
3) 8739

Ax(x-1)(x+2)+Bx(x-1)+Cx+D
= Ax^3 + Ax^2 -2Ax + Bx^2 – Bx + Cx + D
= Ax^3 + (A+B)x^2 + (C-2A-B)x + D

By matching the coefficients,
A = 5
A+B = 3, Hence B=3-5=-2
C-2A-B=-10, Hence C = -10+2A+B= -10 + 2*5 -2 = -2
D = 6

2) 54(46)
= (50+4)(50-4) ……. (using the formula (a+b)(a-b) = a^2 – b^2 )
= 50*50 – 4*4
= 2500 – 16
= 2484

3) 102^2 - 16(108) + 63
= 102^2 – 16(102+6) + 63
= 102^2 – 16*102 + 63 – 6*16
= (102-2)(102-14) -28 + 63 – 96 …. (using the formula (a-2)(a-14)=a^2 -16a +28)
= 100*88 -61
= 8800 – 61
= 8739

Ax(x-1)(x+2) + Bx(x-1) + Cx + D = 5x* + 3X^ - 10x + 6
Ax(x^ + x -2) + B(x^ - x) + Cx + D = 5x* + 3X^ - 10x + 6
A(x* + x^ - 2x) + B(x^ - x) + Cx + D = 5x* + 3X^ - 10x + 6
Ax* + (A + B)x^ + (-2A - B + C)x + D = 5x* + 3X^ - 10x + 6

By comparing the coefficients,

A = 5
A + B = 3, therefore B = -2
-2A - B + C = -10, therefore C = -2
D = 6

54(46)
=(50 + 4)(50 - 4)
=50^ - 4^
=2500 - 16
=2484

102^ - 16(108) + 63
=(3 x 34)^ - 16(108) + 63
=9(34^) - 9[16(12)] + 9(7)
=9[34^ - 16(12) + 7]
=9[34^ - 16^ + 7 + 16(4)]
=9[(34 + 16)(34 - 16) + 7 + 16(4)]
=9[50(18) + 7 + 64]
=9(971)
=9710 - 971
=8739

2007-01-07 17:13:44 補充：
The first person's answer to question 3 has a mistake.(100 - 6)^ = 100^ - 2(100)(6) + 36The last term of the expansion should be +36, not -36.

2007-01-07 17:14:17 補充：
The symbols are the same as the question given.