搜搜考试网1-2y-y-11 3
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/06 07:53:43
(1).（1-y）^2(1+y)(-1-y)=-（1-y）^2(1+y)^2=-[（1-y）(1+y)]^2=-(1-y^2)^2(2).x^2/2+y^2/2=(x^2+y^2)/2=[(x+y)^
∵3y^2-(2y+1)(y-2)=(y-5)(y-1)==>3y^2-2y^2+3y+2=y^2-6y+5==>9y==3==>y=1/3∴原方程的解是y=1/3.
原式=x^2-xy-2xy-2y^2=2x^2-3xy-2y^22、是的,你算错了,以后要细心!原式=(a^2+4ab+4b^2)(2a+b^2)=2a^3+a^2b^2+8a^2b+4ab^3+8a
1、4(x-y+1)+y(y-2x)=4x-4y+4+y²-2xy=y²-4y+4-2xy+4x=(y-2)²-2x(y-2)=(y-2x-2)(y-2)2、4x^4-1
1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)（2a+b)(2
[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y
2y+8y-1=2(y+4y+4)-1-8=2(y+2)-9=0所以(y+2)=2/9即y+2=√2/3或者-√2/3所以y=√2/3-2或者y=-√2/3-2
y-[3y-1]=5【y+2]y-3y+1=5y+103y=-9y=-3
4（x-y+1)+y(y-2x)=4x-4y+4+y^2-2xy=(y^2-4y+4)-2x(2-y)=(y-2)^2+2x(y-2)=(y-2)(2x+y-2)
1)4xy2)-1/5y^23)2ab^2c4)(2a+b)^25)2yz+xz6)-3a-57)4xy
4(x-y+1)+y(y-2x)=4x-4y+4+y^2-2xy=(2-y)^2+2x(2-y)=(2-y)(2-y+2x)4x^4-13x²+9=4x^4-12x^2+9-x^2=(2x^
（1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/
设y'=p,y"=p(dp/dy)y·y''=1+y'^2yp(dp/dy)=1+p^2pdp/(1+p^2)=dy/y(1/2)ln(1+p^2)=ln|y|+c1+p^2=c1y^2p^2=c1y
不显含x型.令y'=p,则y"=pdp/dy,原微分方程可化为yp[dp/dy]+1=p^2即ydp/dy=(p^2-1)/p分离变量p/(p^2-1)dp=dy/y两边积分∫p/(p^2-1)dp=
∵3y^2-(2y+1)(y-2)=(y-5)(y-1)==>3y^2-2y^2+3y+2=y^2-6y+5==>9y==3==>y=1/3∴原方程的解是y=1/3.希望对你有所帮助,
y=C2-ln[cos[x+C1]]dy'/dx=1+(y')^2dy/(1+(y')^2)=dxArcTan(y')=x+C1y'=Tan(x+C1)dy=Tan(x+C1)dxy=C2-ln[co
令y'=p,那么y"=dp/dx=dp/dy*dy/dx=p*dp/dy所以原方程可以化为p*dp/dy+p=py即dp=(y-1)*dy等式两边积分得到p=y'=0.5y^2-y+C(C为常数)x=