x y 2=2x-y 3=x 2
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设x2=y3=z5=k,则x=2k,y=3k,z=5k,∴6k+6k-5k=14,k=2,∴x=4,y=6,z=10.答:x,y,z的值分别为4,6,10.
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
(1)方程组整理得:x+y=8①3x+2y=24②,②-①×2得:x=8,将x=8代入①得:y=0,则方程组的解为x=8y=0;(2)x+y+z=12①x+2y+5z=22②x=4y③,将③代入①得:
原方程组可化为:4x−y=5 ①3x+2y=12②,①×2+②得11x=22,∴x=2,把x=2代入①得:y=3,∴方程组的解为x=2y=3.
x4+y2x2+y4=x^4+2y^2x^2+y^4-x^2y^2=(x^2+y^2)^2--x^2y^2=(x^2+y^2+xy)(x^2+y^2-xy)x3+x2y-xy2-y3=(x-y)(x^
(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的
3xy(x²y-xy²+xy)-xy²(2x²-3xy+2x)=3x³y²-3x²y³+3x²y²-
原式=x3+3x2y-5xy2+6x3+1-2x3+y3+2xy2+x2y+2-4x2y-7x3-y3+4xy2+1=-2x3+xy2+4,由于y为偶次幂,故误把“x=3,y=-1”写成“x=3,y=
原方程可化为:3x+2y=12①2x+3y=28②,①×2-②×3得,-5y=-60,解得y=12,代入①得,3x+24=12,解得x=-4,故此方程组的解为:x=−4y=12.
因为x^2y+xy^2=30,xy(x+y)=30,xy=6x+y=5所以(x+y)^2=x^2+y^2+2xy=x^2+y^2+12=25所以x^2+y^2=13所以(x-y)^2=x^2+y^2-
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²
(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
3x³y-3x²+xy²-2y³
∵x2=y3=z4,∴6x=4y=3z,∵3x-2y+5z=-20,∴6x-4y+10z=-40,∴z=-4,∴x=-2,y=-3,∴x+3y-z=-2+3×(-3)-(-4)=-7;故答案为:-7.
设x2=y3=z4=k,则x=2k,y=3k,z=4k,∴4x−3y+5z2x+3y=4×2k−3×3k+5×4k2×2k+3×3k=1913.故答案为:1913.
设x2=y3=z5=t,则x=2t,y=3t,z=5t,∴x+3y−zx−3y+z=2t+9t−5t2t−9t+5t=-3.故答案是:-3.再问:=v=O(∩_∩)O谢谢
根据题意,设x=2k,y=3k,z=4k∵x+y+z=18∴2k+3k+4k=18,解得k=2∴x=4,y=6,z=8∴x+y-z=2.
条件应为x与y不等.这样的话,x²-y²=x³-y³(x+y)(x-y)=(x-y)(x²+xy+y²)∵x-y≠0∴x+y=x²
x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)
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