dy dx x^2 y=1

[(2x+y)^2+(y+2x)(y-2x)-2y(4x-y)]/4y=(4x^2+4xy+y^2+y^2-4x^2-8xy+2y^2)/4y=(-4xy+4y^2)/4y=-x+y=-1/2+1/3

代数式4y²-2y+5=7,那么代数式2y²-y+1=2∵4y²-2y+5=7∴4y²-2y=22(2y²-y)=22y²-y=1∴2y&#

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

∵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.

再答：诚邀您加入百度知道团队“驾驭世界的数学”。

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+y^-1=3,y可以做分母,y≠0,所以两边可以同时乘以yy^2-3y+1=0,用求根公式得y=（3±√5）/2y+y^-2=y+y^-1×y^-1当y=（3+√5）/2时,y+y^-2=5-√5

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

(2x+1)^2+y^2+2y+1=0(2x+1)^2+(y+1)^2=0(2x+1)^2=0,(y+1)^2=0x=-1/2,y=-1{(x^2+y^2)-(x-y)^2+2y(x-y)}/(2y)

(1-3y)^2+(2y-1)^2=13(y-1)(y+1)1+9y²-6y+4y²+1-4y=13y²-13;10y=15;y=3/2;很高兴为您解答,skyhunte

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

5/y^2+y-1/y^2-y=0两边同时乘以y(y+1)(y-1),得5(y-1)-(y+1)=0解得y=3/2检验：将y=3/2代入原方程,左边=右边所以y=3/2是原方程的解.

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=

题目有误（1）若是(3y+5)/(3y-6)=1/2+(5y-4)/(2y-4),(3y+5)/(3y-6)=1/2+(5y-4)/(2y-4)两边同时乘以12(y-2),得(3y+5)x4=12(y

∴y²+y=2∴y²+y+1/4=2+1/4∴（y+1/2）²=4/9∴y+1/2=±2/3∴y=1或－2再问：用换元法解方程x^+x+1=2/x^+x时，若设x^+x=