中二數學 HELP!!

(a).已知 a^2+b^2=39 and ab=18,找出(a+b)^2 ,(a-b)^2
(a+b)^2
= a^2 + b^2 + 2ab
= 39 + 2(18)
= 75
(a-b)^2
= a^2 + b^2 - 2ab
= 39 - 2(18)
= 3

(b).擴張 (a+b)^2(a-b)^2
= [(a+b)(a-b)]^2
= (a^2 - b^2)^2
= a^4 - 2a^2*b^2 + b^4

(c).根據(a),(b)的結果,找出 a^4-2a^2*b^2+b^4的數值
a^4 - 2a^2*b^2 + b^4
= (a+b)^2(a-b)^2
= (75)(3)
= 225
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2.ABC 是直角三角形.If AB=x BC=y and CA=z, x,y,z的關係是 x^2=(z+y)(z-y). 找出 x, z 的數值 if y=5 z=x+1
x^2 = (z+y)(z-y)
(5 z - 1)^2 = (z+5 z)(z-5 z)
(5 z - 1)^2 = (6 z)(-4 z) = -24z^2
25 z^2 - 10z + 1 = -24z^2
49 z^2 - 10z + 1 = 0
係咪打錯野,有問題..
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3.找出 A,B的數值 (A+2)x+(A-B)恆等於2x+5
(A+2)x+(A-B) = 2x+5
A+2 = 2, A = 0
A-B = 5, B = -5
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4.找出A,B,C的數值 A(x-1)(x+2)+x恆等於2B(x+2)-C
A(x-1)(x+2)+x = 2B(x+2)-C
put x = -2, -2 = -C, C = 2
put x = 1, 1 = 6B - C, B = 1/2
put x = 0, -2A = 4B - C = 0, A = 0
A = 0, B = 1/2, C = 2
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5.找出 A,B,C的數值 A(x^2+x+1)-B(x+2)+4恆等於3x^2+4x+C
A(x^2+x+1)-B(x+2)+4 = 3x^2+4x+C
x = -2, 3A + 4 = 4 + C, 3A = C --------------- (1)
x = 0, A - 2B + 4 = C ----------------- (2)
x = -1, A - B + 4 = -1 + C, A - B + 5 = C ---------- (3)
(3) - (2), B + 1 = 0, B = -1 ----------- (6)
sub (6) into (2), A + 6 = C ------------- (7)
from (1), (7), A + 6 = 3A, A = 3
C = 9
A = 3, B = -1, C = 9
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6.化簡3(x+2)^2-12
= 3 [(x+2)^2-4]
= 3 [(x+2)^2-2^2]
= 3 [(x+2)-2][(x+2)+2]
= 3x(x+4)
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7.化簡 9x^2-4y^2+4y-1
= 9x^2-(4y^2-4y+1)
= 9x^2-(2y-1)^2
= (3x)^2-(2y-1)^2
= [(3x)-(2y-1)] [(3x)+(2y-1)]
= (3x-2y+1) (3x+2y-1)
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8.化簡 4x^2+12x-135
= (2x - 9)(2x + 15)
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9.If p=a+b and q=ab,express a^2+b^2 in terms of p and q
a^2+b^2
= (a+b)^2 - 2ab
= p^2 - 2q