math f.2 急

Q1)
3(x2-1)-Ax ≡ 3x(Bx+1)-C
3x2 - Ax - 3 ≡ 3Bx2 + 3x - C

By equating the coefficients;
3 = 3B .............(1)
-A = 3 ..............(2)
-3 = -C .............(3)

Therefore,
A = -3
B = 1
C = 3


Q2)
2M3 - M2
= M2 (2M - 1)


Q3)
( X-Y )( X2 - XY + Y2 )
= X3 - (X2)(Y) + (X)(Y2) - (X2)(Y) + (X)(Y2) - Y3
= X3 - 2X2Y + 2XY2 - Y3


Q4)
(x-y)2 + 2xy
= x2 - 2xy + y2 + 2xy
= x2 + y2


Q5) 1 / 4(6X+10Y)2
= 1 / 4( 36X2 + 120XY + 100Y2 )
= 1 / (144X2 + 480XY + 400Y2)

OR (1/4)(6X+10Y)2
= ( 1/4 )( 36X2 + 120XY + 100Y2 )
= 9X2 + 30XY + 25Y2

1. L.H.S.= 3(x^2-1) - Ax
= 3x^2 - 3 - Ax

R.H.S.= 3x(Bx+1) - C
=3Bx^2 - C+ 3x

By comparing L.H.S and R.H.S,
A = -3, B = 1, C = 3.

2. 2M^3 - M^2
=M^2(2M-1)

3.(x-y)(x^2-xy+y^2)
=x^3-x^2y+xy^2-x^2y+xy^2-y^3
=x^3 - 2x^2y + 2xy^2 - y^3

4. L.H.S.=x^2 + y^2
R.H.S.= (x-y)^2+2xy
=x^2+y^2
=L.H.S.
∵L.H.S = R.H.S
∴它是恆等式

5.1/4(6X+10Y)^2
= (36x^2 + 120xy + 100y^2)/4
= 9x^2 + 30xy + 25y^2