方程x^2 a^2 y^2 b^2 z^2 c^2=1表示的图形是
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由x+3y-5z=0得x=5z-3y代入2x-y-3z=0中,得10z-6y-y-3z=07z=7y∴y=z代入x=5z-3y中得x=5y-3y=2y∴x=2y于是x∶y∶z=2y∶y∶y=2∶1∶1
∑是循环和例如∑a=a+b+c∑a^2=a^2+b^2+c^2∑(z-y)(x-y)/(x+y-2z)(y+z-2x)=∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z
令x:y:z=1:2:3=kx=k,y=2k,z=3kx+2y+3z=56=k+4k+9k=14kk=4x=4,y=8,z=12再问:已知代数式ax的二次方+bx+c,当x=-2时,它的值为-13:;
x:y:z=1:2:3y=2xz=3xx+2y+3z=x+4x+9x=14x=28x=2y=2x=4z=3x=6
用斯托克斯公式.P=y-z;Q=z-x;R=x-y;原式=二重积分(-1-1)dydz+(-1-1)dzdx+(-1-1)dxdy=-2二重积分(1dydz+1dzdx+1dxdy)=-2*(0+ab
你是要这两个展开么?第一个(a+2b-1)²=[a+(2b-1)]²=a²+2a(2b-1)+(2b-1)²=a²+4ab-2a+4b²-4
e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(
(1)(a+2b/a-b)+(b/b-a)-(2a/a-b);=(a+2b)/(a-b)-b/(a-b)-2a/(a-b)=(a+2b-b-2a)/(a-b)=(-a+b)/(a-b)=-1(2)(y
(1)2x+3y+z=7(2)x+y+z=4(3)3x+y-z=-4(1)和(2)相减得(4)x+2y=3(2)和(3)相加得(5)4x+2y=0(4)和(5)相减:3x=-3;x=-1代入到(4)2
三元一次可知2b-c-3=0a-b=0|a|=0即a=0b=0c=-3再问:答案是a=-1,b=-2,c=-8再答:题目可以打的更准确一点吗再问:就是xyz后面括号里的是次数再答:a-1不等于02b-
①(a-b+c)(-a+b+c)=[c+(a-b)][c-(a-b)=c²-(a-b)²=c²-a²+2ab-b²②(3a+b-2)(3a-b+2)=
A={x|y=√1-x²;,x∈Z}={o,1,-1}B={-1,1,-3}则A∩B={-1,1}以上回答你满意么?
x=2y=3z=13x-y+z=42x+3y-z=12消z得5x+2y=162x+3y-z=12x+y+z=6消z得3x+4y=185x+2y=163x+4y=18合并得x=2y=3代入x+y+z=6
3x-y+z=3(1)2x+y-3z=11(2)x+y+z=12(3)(1)+(2)5x-2z=14(4)(1)+(3)4x+2z=15(5)(4)+(5)9x=29所以x=29/9z=(5x-14)
35a^4b^3/21a^2b^4d=35/21d*a^(4-2)b^(3-4)=5a²/(3bd)x^2-(y-z)^2/(x+y)^2-z^2=(x+y-z)(x-y+z)/[(x+y+
此题的意图在于x+y=?(用换元法)因此此题暗含需要在方程两式x+2y+3z=8;x+3y+5z=6中导出x+y有关的式子;因而涉及到要使用换元法,要在两式中导出x+y的新元;因此两式要乘公因式后相减
3x+2y+z=315①x+2y+3z=285②①+②得5(x+y+z)=600x+y+z=120
x+2y+z=e^(x-y-z)两边对x求偏导注意到z=z(x,y)1+z'=e^(x-y-z)*(1-z')...(1)再对x求偏导z"=e^(x-y-z)(1-z')^2-z"e^(x-y-z).
说x,y,z是正整数吗?再问:对的
1、原式=4a²+4ab+b²-(4a²-b²)=4ab+2b²2、原式=[z+(x+y)][z-(x+y)]+(x-y)²=z²