F2 Maths Help!!!!(40POINTS)!!

1) Determine
x(y-2)+y(z-x)+z(x-y) =0
xy-2x+yz-xy+zx-zy=0
zx – 2x = 0
x(z – 2) = 0
則 x = 0 或 z = 2，y可以是任何數字。

2) Find the constants A&B
A(x-5)+B(x+3) ≡ 4x-4
Ax – 5A + Bx + 3B ≡ 4x – 4
(A+B)x – 5A + 3B ≡ 4x – 4
A + B = 4
及
-5A + 3B = -4
因 A = 4 - B
-5(4 – B) + 3B = -4
-20 + 5B + 3B = -4
8B = 16
B = 2
A + B = 4
A + 2 = 4
B = 2

3) Prove that the following expressions are indentities
(x2+1)(x2-x+1) ≡ x4+x2+1
LHS = (x2+1)(x2-x+1)
= x2(x2-x+1) + (x2-x+1)
= x4 – x3 + x2 + x2 – x + 1
= x4 – x3 + 2x2 – x + 1
不等於 RHS
所以這式是非全等的。

4) Prove that the following expressions are indentities
x(x2+9)(x+3)(x-3) ≡x5-81x
RHS = x5-81x
= x(x4 – 34)
= x(x2 + 32)(x2 – 32)
= x(x2 + 9)(x – 3)(x + 3)
= LHS
所以這式是全等的

5) Expand
1
-- (5b-2a)(5b+2a)
5
= (5b-2a)(5b+2a)/5
= (25b2-4a2) /5
= 5b2 – 4a2/5


(( 3/4) a3 – (2/3) b3)2
= ( 3/4)2 a6 – 2(3/4)(2/3)a3b3 + (2/3) 2 b6
= 9a6/16 - a3b3 + 4b6/9

1.
x(y-2)+y(z-x)+z(x-y) = 0
xy - 2x + yz - xy + xz - yz = 0
- 2x + xz = 0
x( z -2 ) = 0
x = 0 or z =2

2.
A(x-5)+B(x+3) 三 4x - 4
Ax - 5A + Bx + 3B = 4x - 4
(A+B)x - (5A-3B) = 4x - 4

then, you have two equation:
A + B = 4 ... (1)
-5A + 3B = - 4 ... (2)

then put A = 4 - B into equation (2)
By substitution
-5(4 - B) + 3B = -4
-20 + 5B +3B = -4
B = 2

then, A = 4 -B = 4 - 2 = 2

3.
(x^2+1)(x^2-x+1)三 x^4+x^2+1
LHS = (x^2+1)(x^2-x+1)
= x^4 - x^3 + x + x^2 - x +1
=x^4 - x^3 +x^2 +1

RHS = x^4+x^2+1

Thus, LHS is not equal to RHS. They are not indentities

4.
x(x^2+9) (x+3)(x-3)三x^5-81x
LHS = (x^3 + 9x) (x^2 - 9) = x^5 - 9x^3 + 9x^3 - 81x = x^5 - 81x
RHS = x^5 - 81x
LHS is equals to RHS, they are identities

5a.
1/5(5b-2a)(5b+2a)
=1/5 (25b^2 - 4a^2)
= 5b^2 - 4/5a^2

5b.
(3/4 a3 - 2/3 b3 )^2
= (1/12)^2(9a^3 - 8b^3)^2
=1/144 (81a^6 - 144a^3b^3 - 64b^6)
= 9/16a^6 - a^3b^3 - 4/9b^6