dy dx=x^2 y(1 x^3)

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dy dx=x^2 y(1 x^3)
化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1

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

2(x+y) 3x+3y=24 x+y/2x x y/2y= 1

由2(X+Y)3X+3Y=24得:2(X+Y)X+Y=8①;(X+Y/2X)XY/2Y=1得:X+Y=4②;由①、②得出Y=8(1-X),进入②知X=4/7;即Y=24/7

设函数y=y(x)由方程ln(x2+y)=x3y+sinx确定,则dydx|

方程两边对x求导得2x+y′x2+y=3x2y+x3y′+cosxy′=2x−(x2+y)(3x2y+cosx)x5+x3y−1由原方程知,x=0时y=1,代入上式得y′|x=0=dydx|x=0=1

(x+2)(x+3) (x-4)(x+1) (y+4)(y-2) (y-5)(y-3)

(x+2)(x+3)=x²+5x+6(x-4)(x+1)=x²-3x-4(y+4)(y-2)=y²+2y-8(y-5)(y-3)=y²-8y+15x²

已知x-y=1,y≠0,求{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y的值.

已知x-y=1,则y=x-1,x=y+1{(x+2y)²+(2x+y)(x+4y)-3(x+y)(x-y)}÷y=(x²+4xy+4y²+2x²+9xy+4y&

计算3x[xy-2x(1/y-x/3]+3y(x²-y²)

原式=3x(xy-2x/y+2x²/3)+3x²y-3y³=3x²y-6x²/y+2x³+3x²y-3y³=2x

1、x(x-y)(x+y)-x(x+y)^2

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

函数,y=3x/(x^2+x+1) ,x

y=3/(x+1/x+1)x+1/x≤-2,所以x+1/x+1≤-1令t=x+1/x+1,则t≤-1,y=3/t值域为[-3,0)再问:你写的我看不大懂再问:一步步写再答:

求微分方程dydx+y=e

这是一阶线性微分方程,其中P(x)=1,Q(x)=e-x∴通解y=e−∫dx(∫e−x•e∫dxdx+C)=e−x(∫e−x•exdx+C)=e−x(x+C).

(1)(x^2/x)-y-x-y

(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/

先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

计算:1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8x^7/(x^8+y^8

1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【(x+y)-(x-y)】/(x-y)(x+y)-2y/(x^2+y^2)-4y^

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1

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

matlab solve函数 xmaxr=solve(dydx,x)

dydx要是等式才行吧.如果是的话,这句话就是求这个等式的根,用r表示x.

函数y=3x/(x^2+x+1) (x

原式可以化为:y*x^2+(y-3)*x+1=0Δ=(y-3)^2-4y≥0解得y≥9或y≤1由于x

1、(-7x^y)(2x^y-3xy^3+xy) 2、((x-y)^6)/((y-x)^3)/(x-y)

1、(-7x^y)(2x^y-3xy^3+xy)=-14x^(2y)+21x^(y+1)y^3-7x^(y+1)y;2、((x-y)^6)/((y-x)^3)/(x-y)=-(x-y)^3/(x-y)

设函数y=y(x)由方程ex+y+cos(xy)=0确定,则dydx

在方程ex+y+cos(xy)=0左右两边同时对x求导,得:ex+y(1+y′)-sin(xy)•(y+xy′)=0,化简求得:y′=dydx=ysin(xy)−ex+yex+y−xsin(xy).

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2