Question

已知a+b=7,ab=5,则a²＋b³，(a-b)²分别为多少

1.已知a+b=7,ab=5,则a²＋b³，(a-b)²分别为 2.若x+y=1，则x³+y³+3xy= 3.(a-2b+c)(a+2b-c)-(a+2b+c)²= 4.x+y=1,x²+y²=2,则x的七次方加y的七次方等于 5.若x-y=2,x²+y²=4，则x的2001次幂加y的2001次幂的值是

Answer 1

1.
a^2+b^2=(a+b)^2-2ab=39
(a-b)^2=(a+b)^2-4ab=29
2.
原式=
(x+y)(x^2-xy+y^2)+3xy=(x+y)^2=1
3.
原式=[a+(c-2b)]*[a-(c-2b)]-[a+(2b+c)]^2=a^2-(c-2b)^2-a^2-2a(2b+c)-(2b+c)^2
=
-2c^2-8b^2-4ab-2ac
4.
x^7+y^7=(x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)
=x^6+y^6+2x^3y^3
+x^2y^2(x^2+y^2)-3x^3y^3-xy(x^4+y^4)
=4-4xy+3x^2y^2-3x^3y^3-xy*(x^2+y^2)^2+2x^3y^3
=4-8xy+3x^2y^2-x^3y^3
因为xy=0.5*[(x+y)^2-x^2-y^2]=-0.5
所以原式=4+4+0.75+0.125=8.875
5.因为x-y=2,x^2+y^2=4
所以(x-y)^2=4=x^2+y^2
所以xy=0
所以
x^2001+y^2001=
±2^2001
唉
补充：
因为x^2-3x+1=0
x^2+1/x^2=3x-1+1/(3x-1)=
[(3x-1)^2+1]/(3x-1)
=(3x^2-9x+2+3x)/(3x-1)=1
x^3+1/x^3=
3x^2-x+1/(3x^2-x)=8x-3+1/(8x-3)
=[(8x-3)^2+1]/(8x-3)=(144x-54)/(8x-3)=18
x^4+1/x^4=(3x-1)^2+1/(3x-1)^2=9x^2-6x+1+1/(3x-1)^2
=21x-8+1/(21x-8)=
[(21x-8)^2+1]/(21x-8)=(987x-376)/(21x-8)=47
两位数：
设个位为b十位为a
因为a≠b
则平方差为
99(a+b)*|a-b|
∵a+b<20,|a-b|<9
∴a+b=11,
a-b=1,或4
∴符合条件的两位数是：56,65