z=ycos(x-2Y)-搜答案-搜搜考试网

(y-z)^2+(z-x)^2+(x-y)^2=(x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2[(y-z)^2-(y+z-2x)^2]+[(z-x)^2-(x+z-2y)^2]+[(

1／x＝p1／y＝q1／z＝rpq+qr+pr=1(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q

(x+y-z)(x-y+z)=

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

解答在图片中

P=cos(x+y^2)+2yQ=2ycos(x+y^2)+3xP'y=-2ysin(x+y^2)+2Q'x=-2ysin(x+y^2)+3添加线段L1：（π,0)到（0,0)注意由L和L1构成的封闭

我去,这什么题啊

1-y/x*cos(y/x)+cos(y/x)dy/dx=0令y/x=u,则dy/dx=u+xdu/dx所以1-ucosu+cosu*(u+xdu/dx)=0cosu*xdu/dx=-1cosudu=

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

有这样的公式：a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明：（x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3（x+y

xsin(y/x)-ycos(y/x)]dx+xcos(y/x)dy=0②解初值xy'-y=xtany/x,y(1)=π/21.令y/x=t,则方程化为(xsint-xtcost)dx+xtc

x+y+z-6=02x+3y-z-12=02x-y-z=0组成方程组再解x=2y=3z=1

求微分方程ycos(y/x)=[(x²/y)sin(y/x)+xcos(y/x)]dy/dx的通解令u=y/x,则y=ux,dy/dx=u+xdu/dx,代入原方程得：uxcosu=[(x/

合并同类项么,很简单的只要你愿意去做左边=cos*x（cos*y+sin*y）+sin*x（cos*y+sin*y)=cos*x+sin*x=1=右边

设a=x-y,b=y-z,-a-b=z-x(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方b^2+a^2+(-a-b)^2=(-a-b-

将(x+y+z)²展开有(x+y+z)²=x²+y²+z²+2xy+2xz+2yz=x²+y²+z²所以2xy+2xz+

=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)

3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.

设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是：(x+3y-z)/(x-3y+z)?若是,如下：(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3

(x+y-z)^2-(x-y+z)^2=?

根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开：得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)合并同类项