x3-y4=1与3x-4y=2用加减消元法
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曲线y=x3+x-2求导可得y′=3x2+1设切点为(a,b)则3a2+1=4,解得a=1或a=-1切点为(1,0)或(-1,-4)与直线4x-y-1=0平行且与曲线y=x3+x-2相切的直线方程是:
原式分解因式得x^3y^3(2x-y)=(xy)^3(2x-y)=8/3.(x^3表示x的3次方)
把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)
4y4,y,4xy,3x²y,3x3次方
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
原方程组可化为4x-3y=12 ①3x-4y=2 ②,①×4-②×3,得7x=42,解得x=6.把x=6代入①,得y=4.所以方程组的解为x=6y=4.
∵x+y=6,xy=4,∴(1)x2+y2=(x+y)2-2xy,=62-2×4,=28;(2)(x-y)2=x2+y2-2xy,=28-2×4,=20;(3)x4+y4=(x2+y2)2-2x2y2
解(15x^4y^4-9x^5y³-3x^6y²)/(-3x²y)²=(15x^4y^4-9x^5y³-3x^6y²)/(3x²y
由(2)得4x=3y=6z,∴x=34y,z=12y;代入(1)得:y=4,代入(2)得:x=3,z=2,方程组的解为x=3y=4z=2.
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
x2+y2=(x+y)2-2xy=14x3+y3=(x2+y2)×(x+y)-xy2-yx2=14×4-xy(x+y)=52……剩下的就是这么个算法,手机党,求个最佳哈
很明显的,不含x型,二阶微分方程,令y'=t,y''=t*t'y*t*t'-t^2=y^4然后t*t'=(t^2)'/2令t^2=s(y/2)*s'-s=y^4然后用公式解即可,解的过程就不用说了吧
题:集合A={x|x3},B={y|y4},求A∩B=?A∪B=?-A∩B=(-∞,-3)∪(4,+∞)A∪B=(-∞,1)∪(3,+∞)题中的y和x都不影响本题,别在这一点绊住了哪里不懂的话请追问理
y^2+3y-1=0把y=0代入-1=0,不成立所以y不等于0两边除以yy+3-1/y=0y-1/y=-3平方y^2-2+1/y^2=9y^2+1/y^2=11平方y^4+2+1/y^4=121y^4
设x1+y1=x2+y2=x3+y3=x4+y4=X5+y5=A,y1=A-x1,y2=A-x2,...,y5=A-x5.yˉ=(y1+y2+...+y5)/5=(A-x1+A-x2+...+A-x5
x4-xy3-x3y-3x2y+3xy2+y4=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-
(x^4-y^4)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)*(x+y)/(x^2+y^2)=(x^2
X^2+y^2=(x+y)^2-2Xy=a^2-2bx3次方+y3次方=(x+y)(X^2+y^2-xy)=a(a^2-2b-b)=a^3-3abx4次方+y4次方=(x^2+y^2)^2-2x^2y
X:自由度n=3,标准化Xi即Xi=Xi/σ,χ2(3)=(X1^2+X2^2+X3^2)/σ^2Y:因为已知均值,故自由度n=4-1=3,同理χ2(3)=((Y1-A)^2+(Y2-A)^2+(Y3
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²