刀客家族约女人35集:求H.O.T的<<Outside Castle>> MV

1.解:设这个相同的根是a,则
aa+4a+m=0且2aa-3a+m=4.
两式相减,得
aa-7a=4,即aa=7a+4.
把aa=7a+4代入aa+4a+m=0,得7a+4+4a+m=0,即m=-11a-4.
由aa-7a=4解得a1=(7+√65)/2,a2=(7-√65)/2.
把a1,a2的值分别代入m=-11a-4得
m=(-85-11√65)/2或m=(-85+11√65)/2.
2.解:(1)2xx+xy-3yy+x+4y-1=2xx+(y+1)x+(-3yy+4y-1)
=2xx+(y+1)x-(3y-1)(y-1)
=[2x+(3y-1)][x-(y-1)]=(2x+3y-1)(x-y+1);
(2)6xx-xy-15yy-5x+21y-6=6xx-(y+5)x-(15yy-21y+6)
=6xx-(y+5)x-(5y-2)(3y-3)=[3x-(5y-2)][2x+(3y+3)]=(3x-5y+2)(2x+3y+3);
(3)-10xx+23xy+5yy-13x-8y+3=-(10xx-23xy-5yy+13x+8y-3)
=-[10xx-(23y-13)x-(5yy-8y+3)]
=-[10xx-(23y-13)x-(5y-3)(y-1)]
=-[5x+(y-1)][2x-(5y-3)]
=-(5x+y-1)(2x-5y+3).
3.解:
(1)（x+1)(x+2)(x-4)(x-5)=40
[(x+1)(x-4)][(x+2)(x-5)]=40,
(x^2-3x-4)(x^2-3x-10)=40,
(x^2-3x-4)[(x^2-3x-4)-6]=40,
(x^2-3x-4)^2-6(x^2-3x-4)-40=0,
(x^2-3x-4-10)(x^2-3x-4+4)=0,
即x^2-3x-14=0,或x^2-3x=0.
解之,得
x1=(3+√65)/2,x2=(3-√65)/2,x3=0,x4=3.
(2) (x+1)^4+(x+3)^4=706 ,
设x+2=y, 则原方程即为
(y-1)^4+(y+1)^4=706,
即(y^2-2y+1)^2+(y^2+2y+1)^2=706,
y^4+4y^2+1-4y^3-4y+2y^2+y^4+4y^2+1+4y^3+4y+2y^2=706,
2y^4+12y^2+2-706=0,
y^4+6y^2-352=0.
y^2=-3+19=16(舍去负值),
所以,y1=4,y2=-4.
所以,x1=4-2=2,x2=-4-2=-6.