急求数学题！好的，正确的重重加分！急~~~~~~！！！本人初一，哥哥姐姐快来帮帮我啊！！~~~~~~~~~~~~~~~~~

(m-n)(m^2+n^2)(m+n)
=(m²-n²)(m²+n²)
=m^4-n^4

(xy+1)^2(xy-1)^2
=(x²y²-1)²
=x^4y^4-2x²y²+1

(-a-b+c)(a-b-c)
=-(a-c+b)(a-c-b)
=-[(a-c)²-b²]
=-(a²-2ac+c²-b²)
=b²-a²-c²+2ac

(a+b+3)(a-b-3)
=a²-(b+3)²
=a²-b²-6b-9

(x+2y-3/2)(x-2y+3/2)
=x²-(2y-3/2)²
=x²-4y²+6y-9/4

已知a+b=2 ab=15 求a^2+b^2
a²+b²
=(a+b)²-2ab
=2²-2*15
=-26

(a+b)^2=4 (a-b)^2=6 求ab，a^2+b^2
(a+b)²=a²+2ab+b²=4
(a-b)²=a²-2ab+b²=36
两式相加除2
a²+b²=20

已知x+y=7 xy=9 求（x-y）^2
(x-y)²
=(x+y)²-4xy
=7²-4*9
=13

已知a+b=-5 ab=3 求(a-b)^2，a^2+b^2，a^4+b^4
(a-b)²=(a+b)²-4ab=25-12=13
a²+b²=(a+b)²-2ab=25-6=19
a^4+b^4=(a²+b²)²-2a²b²=19²-2*9=343

(a+b+c)^2=? (2x-3y+1)^2
(a+b+c)²=a²+b²+c²+2ab+2ac+2bc
(2x-3y+1)²=4x²+9y²+1-12xy+4x-6y

a=2008x+2004 b=2008x+2005 c=2008x+2000
求a^2+b^2+c^2-ab-ac-bc
a²+b²+c²-ab-ac-bc
=[(a-b)²+(a-c)²+(b-c)²]/2
=(1+16+25)/2
=23

已知x+y+z=2 xt+zy+yz=-5
求x^2+y^2+z^2
x²+y²+z²
=(x+y+z)²-2(xy+yz+xz)
=4+10
=14

多项式4x^2+1加一个单项式后，能成为一个整式的完全平方，则加上的单项式可以是
1.4x
2.-4x
3.-1
解方程
3(x+1)(x-1)-3(x-1)^2-10=2
x=3
(2x-1)^2-(1-3x)^2=5(1-x)(x+1)
x=2.5
(x-3)^2+(5+x)^2=2x^2-6
x=-10
(x+5)(x-5)+2(x+1)^2=3x^2-x+2
x=5

(m-n)(m^2+n^2)(m+n)
=（m^2-n^2)(m^2+n^2)
=m^4-n^4

(xy+1)^2(xy-1)^2
=[(xy)^2-1]^2
(xy)^4-2(xy)^2+1

(-a-b+c)(a-b-c)
=-[(a-c)+b][(a-c)-b]
=b^2-(a-c)^2
=b^2-a^2-c^2+2ac

(a+b+3)(a-b-3)
=[a+(b+3)][a-(b+3)]
=a^2-(b+3)^2
=a^2-b^2-6b-9

(x+2y-3/2)(x-2y+3/2)
=[x+(2y-3/2)][x-(2y-3/2)]
=x^2-(2y-3/2)^2
=x^2-4y^2+6y-(9/4)

已知a+b=2 ab=15 求a^2+b^2
解：因为a+b=2
所以：（a+b)^2=4，即a^2+b^2+2ab=4
所以：a^2+b^2=4-2ab=4-30=-26

(a+b)^2=4 (a-b)^2=6 求ab，a^2+b^2
解：由已知得a^2+b^2+2ab=4， a^2+b^2-2ab=6
所以：a^2+b^2=5 ，ab=-1/2

已知x+y=7 xy=9 求（x-y）^2
解：由x+y=7得x^2+y^2+2xy=49
所以：x^2+y^2=49-2xy=49-18=31
所以：（x-y)^2=x^2+y^2-2xy=31-18=13

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