已知a²+b²-4a-6b+13=0求a+b的值（ay+bx）³-（ax+by）³+（a³-b³）（x³-y³）x4+x³+2x²+x+1已知n是正整数,且n4-16n²+100是质数那么n=?（1+y）²-2x²（1+y²）+x4（1-

Question

已知a²+b²-4a-6b+13=0求a+b的值
（ay+bx）³-（ax+by）³+（a³-b³）（x³-y³）
x4+x³+2x²+x+1
已知n是正整数,且n4-16n²+100是质数 那么n=?
（1+y）²-2x²（1+y²）+x4（1-y）²
-a4-b4-c4+2a²b²+2b²c²+2a²c²
x³+3x²-4
x4+2x³-9x²-2x+8
a³+3a²+3a+b³+3b²+3b+2
x²-y²+2x+6y-8
-14x²y²+x4+y4
x4-47x²+1
x4+1/4y4
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Answer 1

a²+b²-4a+6b+13=（a-2）²+（b+3）²=0
∴a=2,b=-3
a+b=-1
（ay+bx）³-（ax+by）³+（a³-b³）（x³-y³）
=（ay+bx-ax-by）[（ay+bx）²+（ay+bx）（ax+by）+（ax+by）²]+（a-b）（a²+ab+b²）（x-y）（x²+xy+y²2）
=（x-y）（b-a）[（a2+b2）（x2+xy+y2）+ab（x2+4xy+y2）]-（b-a）（x-y）[（a2+b2）（x2+xy+y2）+ab（x2+xy+y²）]
=（x-y）（b-a）[ab•3xy],
=3abxy（x-y）（b-a）
x4+x³+2x²+x+1
=x4+2x²+1 +x³+x
=(x²+1)²+x(x²+1)
=(x²+1)(x²+1+x)
=(x²+1)(x+1/2+√3/2)(x+1/2-√3/2)
n4-16n2+100
=n4+20n2+100-36n2
=(n2+10)2-(6n)2
=(n2+10+6n)(n2+10-6n)
n4-16n²+100是质数
又因为n为正整数,
所以n2+10+6n>1.
所以n2+10-6n=1.
所以n=3
（1+y）²-2x²（1+y²）+x4（1-y）²
=(1+y)^2+2(1+y)x^2(1+y)+x^4(1-y)^2-2(1+y)x^2(1-y)-2x^2(1+y^2)
=[(1+y)+x^2(1-y)]^2-2(1+y)x^2(1-y)-2x^2(1+y^2)
=[(1+y)+x^2(1-y)]^2-(2x)^2
=[(1+y)+x^2(1-y)+2x]·[(1+y)+x^2(1-y)-2x]
=(x^2-x^2y+2x+y+1)(x^2-x^2y-2x+y+1)
=[(x+1)^2-y(x^2-1)][(x-1)^2-y(x^2-1)]
=(x+1)(x+1-xy+y)(x-1)(x-1-xy-y)
-a4-b4-c4+2a²b²+2b²c²+2a²c²
=-（a^4+b^4+c^4+2a²b²-2b²c²-2a²c²-4a²b²）
=-【(a^4+b^4+2a²b²)+c^4-2a²c²-2b²c²-4a²b²】
=-【(a²+b²)²+(c²)²-2c²(a²+b²)-4a²b²】
=-【(a²+b²-c²)²-4a²b²】
=-(a²+b²-c²-2ab)(a²+b²-c²+2ab)
=-[(a-b)²-c²][(a+b)²-c²]
=-(a-b+c)(a-b-c)(a+b+c)(a+b-c)
x^3-3x²+4
=(x^3+x²)-4(x²-1)
=x²(x+1)-(x+1)(4x-4)
=(x+1)(x²-4x+4)
=(x+1)(x-2)²
x^4+2x^3-9x^2-2x+8
=x^2(x^2-1)+2x^3-8x^2-2x+8
=x^2(x^2-1)+2x(x^2-1)-8x^2+8
=x^2(x^2-1)+2x(x^2-1)-8(x^2-1)
=(x^2-1)(x^2+2x-8)
=(x+1)(x-1)(x-2)(x+4)
a³+3a²+3a+b³+3b²+3b+2
=a³+3a²+3a+1+b³+3b²+3b+1
=（a+1）³+（b+1）³
x²-y²+2x+6y-8
=x²+2x+1-y²+6y-9
=（x+1）²-(y-3）²
=(x+y-2)(x-y+4)
-14x²y²+x4+y4
=-2x²y²+x4+y4-12x²y²
=(x²-y²)²-12x²y²
=(x²-y²+2√3xy)（x²-y²-2√3xy)
x4-47x²+1
=x4+2x²+1-49x²
=（x²+1）²-49x²
=（x²+9x+1）（x²-9x+1）
x4+1/4y4
=x4+1/4y4+x²y²-x²y²
=(x²+1/2y²)²-x²y²
=(x²+1/2y²+xy)(x²+1/2y²-xy)