有(X^4+X^2-4)(X^4+X^2+3)+10 .设X^4+X^2=y,则原式=(Y-4)(Y+3)+10=Y^2

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/24 03:06:05

有(X^4+X^2-4)(X^4+X^2+3)+10 .设X^4+X^2=y,则原式=(Y-4)(Y+3)+10=Y^2-Y-2=(Y-2)(Y+1)=(X^4+X^2-2)(X^4+X^2+1)=(X^2+2)(X+1)(X-1)(X^4+X^2+1).其中(X^4+X^2-2)(X^4+X^2+1)=(X^2+2)(X+1)(X-1)(X^4+X^2+1)这一步是怎么来的,我实在是没有什么分,只求哪位路过的大侠助一臂之力,

(X^4+X^2-2)=（X^2+2）（x^2-1）=（X^2+2）(x+1)(x-1)
第一步看成关于x^2的因式.

(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20 {(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20 4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6 3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27 设实数X,Y满足2X+Y-2>=0,X-2Y+4>=0,3X-Y (x+y)(x+2y)(x+3y)(x+4y)=-40 已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y 已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-( 数学题……555{(3x+y)(3x-y)-(x-5y)(5x-y)-(x-2y)²÷(-4x),x=-2.y 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1