用待定系数法分解x²-xy-y²-x 5y-2

将2x^2+xy-y^2因式分2x^2+xy-y^2＝(2x-y)(x+y)那么假设2x^2+xy-y^2-4x+5y-6可以分解为(2x-y+a)(x+y+b)展开：2x^2+xy-y^2+(a+2

（以下过程均是在实数范围内分解因式）解(1)x^5+x+1因为原式是5次式所以若原式可以因式分解,则一定可以分解为一个2次式因式和一个3次因式,或者一个1次因式和一个4因式若原式可以分解为一个2次式因

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

x^2+xy-3y^2=x^2+xy+y^2/4-y^2/4-3y^2=(x+y/2)^2-13y^2/4=(x+y/2-√13y/2)(x+y/2+√13y/2)

解2x²y-x³-xy²=x(2xy-x²-y²)=-x(x²-2xy+y²)=-x(x-y)²

：(x^2-xy)+z(x-y)=x（x-y）+z（x-y）=（x-y）（x+z）

设t=xy+1;(xy+1)(x+1)(y+1)+xy=t(x+1)(y+1)+xy=t(xy+x+y+1)+xy=t(t+x+y)+xy=t^2+xt+yt+xy=(t+x)(t+y)-xy+xy=

令x+y=a,xy=b原式化为：（b-1）^2+(a-2)(a-2b)=b^2-2b+1+a^2-2a-2ab+4b=b^2+2b+1+a^2-2a-2ab=(a^2-2ab+b^2)+2(b-a)+

你第一个是不是写错了?[2][(x-1)(y-1)]^2[3](x^3-x+1)(x^2+x+1)

比如Fe2O3+H2SO4==Fe2(SO4)3+H2O设元.xFe2O3+yH2SO4==xFe2(SO4)3+yH2O首先铁,H平衡.得y=3x,所以x=1,y=3其次SO4,O平衡.这个是比较简

是不是x^3y+x^2y+2xy^2-4xy^3原式=xy(x^2-4y^2)+x^2y+2xy^2=xy(x+2y)(x-2y)+xy(x+2y)=xy(x+2y)(x-2y+1)

=(x²-2xy+y²)²=(x-y)的4次方

这个题目看结构,应该是出错了,好像应该是(x^2-2xy)^2+2y^2(x^2-2xy)+y^4只有这样,才能形成完全平方公式(x^2-2xy)^2+2y^2(x^2-2xy)+y^4=(x^2-2

你的式子是无法分解的,已经是最简代数式了x^2-xy+y^2=（x-y/2）^2+3y^2/4恒大于0,无法分解

1.3x2+5xy-2y2+x+9y+n=(3x-y+4)(x+2y+a)=3x2+5xy-2y2+(4+3a)x+(8-a)y+4a∴4+3a=1,8-a=9,a=-1,n=4a=42.ax^2+b

令6x^2+13xy+6y^2+5x+5y+1=(ax+by+c)(dx+ey+f)6x^2+13xy+6y^2+5x+5y+1=(ax+by+c)(dx+ey+f)=adx^2+(ae+bd)xy+

（1）x方+2xy+y方+x+y-2=(x+y)²+(x+y)-2=(x+y+2)(x+y-1)（2）x方+xy-2y方+2x+7y-3=(x+2y)(x-y)+2x+7y-3=(x+2y-

原式=(x+y)(x+y)+2(x+y)xy+(xy)^2-1^2=(x+y)^2+2(x+y)xy+(xy)^2-1=(x+y+xy)^2-1=(x+y+xy-1)(x+y+xy+1)=(x+y+x

（x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

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