1.x+y=a,xy=b 求X2次方+y2次方 x3次方+y3次方 x4次方+y4次方 x5次方+y5次方 2.x+x分之一=3

Question

求x平方+x平方分之一 x立方+x立方分之一 x4次方+x4次方分之一
3.（a-b)(a+b)3次方-2ab(a平方-b平方） (a+b+c)(a+b-c)(a-b+c)(-a+b+c)
4.已知a平方=a+1,求代数式a5次方-5a+2
求解答，20分给你（一体5分，要过程，好的话再加5分！！！）

Answer 1

解：
1、
x+y=a,xy=b
①x²+y²
=x²+2xy+y²-2xy
=(x+y)²-2xy
=a²-2b
②x³+y³
=(x³+3x²y+3xy²+y³)-(3x²y+3xy²)
=(x+y)³-3xy(x+y)
=a³-3ab
③x^4+y^4
=(x^4+2x²y²+y^4)-2x²y²
=(x²+y²)²-2(xy)²
=(a²-2b)²-2b²
=a^4-4a²b+4b²-2b²
=a^4-6a²b+4b²
④
x^5+y^5
=(x²+y²)(x³+y³)-x²y²(x+y)
=(a²-2b)(a³-3ab)-ab²
=a^5-2a³b-3a³b+6ab²-ab²
=a^5-5a³b+5ab²
2、
①
x+1/x=3
两边平方
x²+2x/x+1/x²=3²
x²+2+1/x²=9
x²+1/x²=7
②
x+1/x=3
两边立方
x³+3x²/x+3x/x²+1/x³=3³
x³+3x+3/x+1/x³=27
x³+3(x+1/x)+1/x³=27
x³+1/x³=27-3(x+1/x)
=27-3×3
=27-9
=18
③
x^4+1/x^4
=(x²+1/x²)²-2x²(1/x²)
=7²-2
=47
3、
①
(a-b)(a+b)³-2ab(a²-b²）
=(a-b)(a+b)(a+b)²-2ab(a+b)(a-b)
=(a-b)(a+b)[(a+b)²-2ab]
=(a-b)(a+b)(a²+2ab+b²-2ab)
=(a²-b²)(a²+b²)
=a^4-b^4
②
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)
={[(a+b)+c][(a+b)-c]}[c+(a-b)][c-(a-b)]
=[(a+b)²-c²][c²-(a-b)²]
=-(a²+2ab+b²-c²)(a²-2ab+b²-c²)
=-[(a²+b²-c²)+2ab][(a²+b²-c²)-2ab]
=-[(a²+b²-c²)²-(2ab)²]
=-[(a²+b²-c²)²-4a²b²]
=-[(a²+b²)²+2(a²+b²)c²+c^4-4a²b²]
=-[a^4+2a²b²+b^4-2(a²+b²)c²+c^4-4a²b²]
=-[a^4-2a²b²+b^4-2(a²+b²)c²+c^4]
=-a^4-b^4-c^4+2a²b²+2a²c²+2b²c²
4、
解法1：
a²=a+1
两边平方
a^4=(a²)²=(a+1)²=a²+2a+1=(a+1)+2a+1=3a+2
a^5=aa^4=a(3a+2)=3a²+2a=3(a+1)+2a=5a+3
所以
a^5-5a+2
=(5a+3)-5a+2
=5
解法2：
a²=a+1
采用逐渐降次法
a^5-5a+2
=a²a²a-5a+2
=(a+1)(a+1)a-5a+2
=(a+1)²a-5a+2
=(a²+2a+1)a-5a+2
=[(a+1)+2a+1]a-5a+2
=(3a+2)a-5a+2
=3a²+2a-5a+2
=3a²-3a+2
=3(a+1)-3a+2
=3a+3-3a+2
=5

Answer 2

1，因为：x+y=a xy=b
所以：
x^2+y^2=(x+y)^2-2xy=a^2-2b
x^3+y^3=(x+y)(x^2+y^2-xy)=a(a^2-2b-b)=a^3-3ab
x^4+y^4=(x^2+y^2)^2-2x^2y^2=(a^2-2b)^2-2(b)^2=a^4-4a^2b+4b^2-2b^2=a^4-4a^2b+2b^2
x^5+y^5=(x^4+y^4)(x+y)-xy(x^3+y^3)=(a^4-4a^2b+2b^2)(a+b)-b(a^3-3ab)=a^5+a^4b-5a^3b-4a^2b^b+5ab^2+2b^3
2,因为x+1/x=3
所以
x^2+1/x^2=(x+1/x)^2-2=3^2-2=7
x^3+1/x^3=(x+1/x)(x^2+1/x^2-1)=3(7-1)=18
x^4+1/x^4=(x^2+1/x^2)^2-2=7^2-2=47
3,(a-b)(a+b)^3-2ab(a^2-b^2)
=(a^2-b^2)(a+b)^2-2ab(a^2-b^2)
=(a^2-b^2)[(a+b)^2-2ab)]
=(a^2-b^2)(a^2+b^2)
=a^4-b^4
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)
=[(b+c)^2-a^2][a^2-(b-c)^2]
=a^2(b+c)^2-a^4-(b^2-c^2)^2+a^2(b-c)^2
=a^2(2b^2+2c^2)-a^4-b^4-c^4+2b^2c^2
=2a^2b^2+2a^2c^2-a^4-b^4-c^4+2b^2c^2
=-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2
4,因为a^2=a+1
所以
a^2+a=1
a^5=a^4+a^3
a^4=a^3+a^2
a^3=a^3+a
a^5-5a+2
=a^4+a^3-5a+2
=a^3+a^2+a^3-5a+2
=2a^3+a^2-5a+2
=2(a^2+a)+a^2-5a+2
=2a^2+2a+a^2-5a+2
=3a^2-3a+2
=3(a^a-a)+2
=3*1+2
=5

Answer 3

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