3xy2分之x2-4y2×x+2y分之xy
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设2分之x=3分之y=4分之z=k∴x=2k,y=3k,z=4kx2+y2+z2分之xy+yz+zx=(6k²+12k²+8k²)/(4k²+9k²+
x+y+xy=9x+y=9-xyx^2y+xy^2=20xy(x+y)=20xy(9-xy)=20xy^2-9xy+20=0(xy-4)(xy-5)=0xy=4或xy=5x+y=5或x+y=4x^2+
由已知:xy+x+y=17,xy(x+y)=66,可知xy和x+y是方程t2-17t+66=0的两个实数根,得:t1=6,t2=11.即xy=6,x+y=11,或xy=11,x+y=6.x2+y2=(
∵数与字母乘积的代数式叫做单项式,∴③0,④-xy2是单项式;∵几个单项式的和是多项式,∴①3x+5y ②x2+2x+y2是多项式.故填空答案:2,2.
①x2y+xy2=xy(x+y)=1×3=3;②x2+y2=(x+y)2-2xy=32-2×1=7.
X2+Y2+8X+6Y+25=0x²+8x+16+y²+6y+9=0(x+4)²+(y+3)²=0∴x+4=0y+3=0x=-4y=-3X2+4XY+4Y2分之
已知x=2-√3,y=2+√3,求代数式x2次方+y2次方值xy=4-3=1x+y=4x²+y²=(x+y)²-2xy=16-2=14再问:谢谢你,但我要那种初级复杂的过
∵x+y=0,xy=-7,∴①x2y+xy2=xy(x+y)=-7×0=0;②x2+y2=(x+y)2-2xy=14.
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
3xy(x²y-xy²+xy)-xy²(2x²-3xy+2x)=3x³y²-3x²y³+3x²y²-
若是209,则xy=8,x+y=15,算出x,y就不是整数了,与题意不符.若是34,x,y为3,5,符合题意.
1设Z=cos(xy2)+3x/x2+y2,计算δz/δyδz/δy=-2xy*sin(xy2)-(3x*2y)/(x2+y2)22、设Z=f(x2-y2,exy),其中f(u,v)为可微函数,求dz
答案是11分之14. 求采纳
即(x²+4x+4)+(y²-6y+9)=0(x+2)²+(y-3)²=0所以x+2=y-3=0x=-2,y=3所以原式=(4+4)/(4+27)=8/31
x2-9a2+12a-4=x2-[(3a)2-2*3a*2+4]=x2-(3a-2)^2=(x+3a-2)(x-3a+2)x2y+3xy2-x-3y=xy(x+3y)-(x+3y)=(xy-1)(x+
x²+4y²-4x+4y+5=0(x-2)²+(2y+1)²=0x-2=0x=22y+1=0y=-1/2x-y=2+1/2=5/2x²y-xy
--2分之1(xy--x^2)+3(y^2--2分之1x^2)+2(4分之1xy--2分之1y^2)=--xy/2+x^2/2+3y^2--3x^2/2+xy/2--y^2=--x^2+2y^2,当x
后面的数字是次方?这样子输入很容易错误.加我Q慢慢帮你解答.
x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup
解;∵x+y=0,xy=-7∴x2y+xy2=xy(x+y)=-7×0=0x2+y2=(x+y)2-2xy=02-2×(-7)=0+14=14.