2x y=7 2x 5y=19

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若x+y为有理数,且|x+1|+(2x-y+4)2=0,则代数式x5y+xy5=______.

根据题意得,x+1=0,2x-y+4=0,解得x=-1,y=2,∴x5y+xy5=(-1)5×2+(-1)×25=-2-32=-34.故答案为:-34.

先化简,再求值 ⒈2(Xy+Xy)-3(Xy-xy)-4Xy,其中X=1,y=-1

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

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

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

xy'=y+xy的

xdy=(y+xy)dxdy/y=((1+x)/x)dxln|y|=ln|x|+x+cy=±e^(ln|x|+x+c)其中c是常数再问:真还不理解我们是选择题:y=cxe^xy=c+x-x^2y=cs

已知(x+1)²+|y-1|=0,求2(xy-5xy²)-(3xy²-xy)得值

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

当x=3,y=3分之1时,求代数出3xy-[2xy-2(xy-2分之3xy)+xy]+3xy的值

3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2

z=sin(xy)+cos^2(xy)一阶偏导数

∂Z/∂x=y*cos(xy)-2cos(xy)*sin(xy)*y=y*cos(xy)-y*sin(2xy)∂Z/∂y=x*cos(xy)-2cos(

已知:xy+x=-1,xy-y=-2.

(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

已知xx+xy=1,xy-yy=-4,则xx+2xy-yy=

xx+2xy-yy=-3两式相加即可

由已知x+y=-2,xy=3那么2(x+xy)-[(xy-3y)-x]-(-xy)等于多少?

2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3*(-2)+2*3=0

化简:xy分之3x^2+2xy-xy分之2x^2-xy=

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

二元二次方程求解xy^2=23400xy=1800

xy^2=xy×y=23400把已知的xy=1800代入上面的公式,求得y=13,再把y=13代入xy=1800,求得x=1800/13

设e^xy-xy^2=Siny,求dy/dx

你好!两边对x求导:e^(xy)*(y+xy')-y^2=y'cosy解得y'=(y^2-ye^(xy))/(xe^(xy)-cosy)

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

求dx/dy-3xy=xy^2的通解

dx/dy-3xy=xy^2dx/x=(y^2+3y)dy两边积分得:lnx=y^3/3+3y^2/2+c==>x=exp(y^3/3+3y^2/2+c)=Cexp(y^3/3+3y^2/2)C常数

已知x5y ……(1) 两边都减5,得0>5y-5x……(2) 即

错在第(4)步.∵x>y,∴y-x<0.不等式两边同时除以负数y-x,不等号应改变方向才能成立.

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy),其中x=20,y=-25分之1

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+

已知:x+y=1,xy=-3,求下列各式的值:(1)x2+y2; (2)x3+y3; (3)x5y+xy5.

再问:能把第三题重新发一遍吗?再答:这三个题本质上式连在一起的再答:这道题应该是希望杯的试题

求通解,dy/dx-3xy=xy^2

dy/dx=xy²+3xydy/dx=x(y²+3y)∫1/[y(y+3)]dy=∫xdx(1/3)∫(3+y-y)/[y(y+3)]dy=∫xdx∫[1/y-1/(y+3)]dy