求出平面x=0,y=0,x y=1所围成的柱体被平面z=0及抛物面-搜答案-拍题作业网

原式=(y-x)/xy=-xy/xy=-1

移项,x=y+xy:同除y得x/y=1+x;同除X得1/y=1/x+1;移项1/x-1/y=-1

y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5

1、x+y=4,可知y=4-x,带入x²-y²+8y,得：x²-(4-x)²+8(4-x)=16；2、x³y+xy³=xy(x²+

由题设可知y=xy-1，∴x=yx3y=x4y-1，∴4y-1=1，故y=12，∴12x=x，解得x=4，于是x+y=4+12=92．故答案为：92．

x+3xy-y/2x-xy-2y=3xy+x-y/2x-2y-xy=6xy/2(x-y)-xy=6xy/6xy-xy=6xy/5xy=6/5

即(3x+y)(x-3y)+10x+10y+k=0是两条直线则可以分解为两个一次式的积(3x+y+a)(x-3y+b)=0由x,y系数和常数3b+a=10b-3a=10ab=kb=4,a=-2k=-8

由对称性知圆X^2+(Y-R)^2=R^2y'=-X/(Y-R)Y''=-1/(Y-R)-X^2/(Y-R)^3把（0,0）代入y''=1/Ry=x^2在（0,0）的二阶导数为2所以R=0.5X^2+

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5

1.在坐标轴上2.在第一和第三象限3.在直线y=-x上

题目错了XY=10再问：不好意思，题目打错了。为什么可以求出xy=10？再答：lgx+lgy=lgxy=1=lg10xy=10

x=0:0.1:100;%假设步长为0.1y=x.*sin(x);ind_peak=intersect(find(diff(y)>0)+1,find(diff(y)

xy+1/xy>=2√（xy*1/xy）=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√（x/y*y/x）=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

解x平方+y平方+3/2xy=(x+y)平方-2xy+3/2xy=(x+y)平方-1/2xy=6平方-1/2×8=36-4=32

∫∫√(y²-xy)dxdy=∫dy∫√(y²-xy)dx=∫dy∫√(y²-xy)(-1/y)d(y²-xy)=∫{(-1/y)(2/3)[(y²-

xy>0得x,y同号又因为x+y

x+y=5xy（2x-3xy+2y)/(x+2xy+y）=[2（x+y）-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问：若x＋1/x＝3,求(x

因为,x-y=3xy所以：-3x+6xy+3y/[(7x-5y)-9xy-(3x-y)]=[-3(x-y)+6xy]/[(7x-5y)-9xy-(3x-y)]=[-3*3xy+6xy]/[7x-5y-

Ix-2I+(y+1)²=0,绝对值与平方均为非负数,则有x=2,y=-15x²+3xy-5y²-2xy+4y²-4x²=x^2-y^2+xy=4-1