已知x 2y=0,求分式x^2-xy分之2xy y^2
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x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
由已知1,分母为0,b=6,由已知2,a=-5.则6a+9b=6*6+9*(-5)=-9
x=-2y(2xy+y^2)/(x^2-xy)=(-4y^2+y^2)/(4y^2+2y^2)=-3y^2/(6y^2)=-1/2再问:我算出来也是-1/2,可答案说是-15/28
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
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)=(
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3
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|=
|x-2|+(y+3)²=0都是非负式所以分别都=0所以x-2=0y+3=0所以x=2y=-3又因为z是最大的负整数所以z=-1原式=2(x²y+xyz)-3(x²y-x
(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)/(2y)=X^2-y^2-y^2+X^2+y^2-2xy-3x^2+xy=-x^2-y^2-xy=-(x^2+y^2+xy-3
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
X=-2时,分式无意义,所以a-2b+3×(-2)=0,即a-2b=6x=1时,分式的值为0,所以1+a+b=0,即a+b=-1a、b的两个方程联立解得:a=4/3,b=-7/3
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方
由题意得(x-2)平方+(y-2)平方+(x-y)平方=0,故x=y=2,故x平方y=8
由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6
xy+x+y=23,x²y+xy²=120,xy(x+y)=120把xy,x+y看成是z²-23z+120=0的两根解得z1=15,z2=8又把x,y看成是m²
∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a
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