y-2y y=1 x x^2通解
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(x-y)(x+y)(xx-yy)=(x^2-y^2)(x^2-y^2)=x^4-2x^2y^2+y^4
X:Y:Z=1:2:3因为:14(XX+YY+ZZ)=(X+2Y+3Z)^214(XX+YY+ZZ)-(X+2Y+3Z)^2=013X^2+10Y^2+5Z^2-4XY-6XZ-12YZ=0(4X^2
(x²+xy-12)²+(xy-2y²-1)²=0由于平方数都大于或等于0,所以上式成立的前提是:(x²+xy-12)²=0,即:x&sup
x=1/(√3-2)=-(2+√3)y=1/(√3+2)=2-√3x+y=-2√3xy=-1xx+xy+yy=(x+y)^2-xy=12-(-1)=13(xx+xy+yy)/(x+y)=-13/2√3
xx+2xy-yy=-3两式相加即可
a+b=(y-x)+(2x-y)=x2a+b=(2y-2x)+(2x-y)=y.(2)cost=x.y/(|x||y|)由a⊥b知a.b=0.x.y=(a+b).(2a+b)=a.2a+b.b=2+1
是2个构造函数.一个默认的不带参.一个带参.带参的接受2个参数.在调用setPoint给x,y赋值再问:为什么方法serPoint是一个参数?而构造方法确实俩个参数?(XX,YY)再答:publicv
2/9再问:过程,谢谢再答:由题目得y/x=2/3xy/xx+yy-yy/xx-yy=y/x-(y/x)²=2/3-4/9=2/9
x2+y2-z2+2xy/x2-y2+z2-2xz=(x+y)2-z2/(x-z)2-y2=(x+y-z)(x+y+z)/(x-y-z)(x-z+y)=(x+y+z)/(x-y-z)然后就是代入了
原式=[(x²+y²-2y²)/y]/[(x-y)/xy]×1/(x²-y²)=[(x²-y²)/y]×[xy/(x-y)]×1/
xx+yy+4x-6y+13=0整理得:(x+2)^2+(y-3)^2=0那么只有(x+2)=0(y-3)=0x=-2y=3(x^2-2x)/(x^2+3y^2)=(4+4)/(4+3*9)=8/31
x^2+x-x^2-y=3x-y=3(x-y)^2=9x^2+y^2-2xy=9(x^2+y^2)/2-xy=9/2
令a=XX+YY则(a-2)a=3a²-2a-3=0(a-3)(a+1)=0a=3,a=-1因为XX+YY≥0所以XX+YY=3
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
(xx)+(yy)-4x+6y+13=0(xx)-4x+(yy)+6y+13=0(xx)-4x+4+(yy)+6y+9=0(x-2)²+(y+3)²=0x-2=0y+3=0x=2y
x^2-2x+y^2+6y+10=0(x-1)^2+(y+3)^2=0所以x=1,y=-3x+y=-2x^2表示x的平方
x(x+1)-(xx+y)=-3x^2+x-x^2-y=-3x-y=-3(xx+yy)/2-xy=(x^2+y^2-2xy)/2=(x-y)^2/2=(-3)^2/2=9/2再问:是对的吧!再答:当然
方程组:X-Y=2……①X^2-Y^2=12……②②÷①得X+Y=6……③①+③得:2X=8,X=4,把X=4代入③得:Y=2,X=4Y=2