請教簡單的方程式問題

1.
用計算機必定是最快,但如果不用計算機的話,用二項式定理
a^4-b^4
=(3+√2)^4 - (3-√2)^4
=3^4 + 4(3^3)(√2) + 6(3^2)(√2^2) + 4(3)(√2^3) + √2^4 - (3^4 - 4(3^3)(√2) + 6(3^2)(√2^2) - 4(3)(√2^3) + √2^4)
= 2(4)(3^3)(√2) + 2(4)(3)(√2^3)
=2(4)(3^3)(√2) + 4(4)(3)(√2)
=264√2

2.
(3+√2)a+(1-2√2)b=3+2√2
3a + √2a + b -2√2b = 3 + 2√2
(3a + b) + (a - 2b)√2 = 3+2√2

3a + b = 3------------------(1)
a - 2b = 2
∴a = 2b + 2---------------(2)
將(2)代入(1):
3(2b+2) + b = 3
7b + 6 = 3
b= -3/7

將 b=-3/7 代入(2):
a= 2(-3/7) + 2
=8/7

3(1)
ab=2---------------------------(1)
a+b=4
∴a=4-b-----------------------(2)
將(2)代入(1):
(4-b)b = 2
b^2 - 4b + 2 = 0
b = {4±√[4^2 - 4(1)(2)]}÷2
b = 2 ±√2
當a=2 +√2,b=2 -√2
當a=2 -√2,b=2 +√2

(2)
a/b+b/a
=(a^2 + b^2)/ab
=[(a+b)^2 - 2ab]/ab
=[4^2 - 2(2)]/2
=6

(3)
ab+√2 a-2b-2√2
=a(b +√2) - 2(b+√2 )
=(a - 2)(b +√2 )
當a=2 +√2,b=2 -√2時,
ab+√2 a-2b-2√2
=(2 +√2 - 2)(2-√2 +√2 )
=2√2
當a=2 -√2,b=2 +√2時,
ab+√2 a-2b-2√2
=(2 -√2 - 2)(2 +√2 +√2 )
= -√2(2 + 2√2)
=-4 - 2√2

4.
(ax+b)^2=x^2-px+4
a^2 x^2 + 2abx + b^2 = x^2 - px + 4
∴a^2 = 1
a = -1 或 1(捨去)

b^2 = 4
b = 2 或 -2(捨去)

-p = 2ab
p = -2(2)(-1)
p = 4