x 3y=2 y 3z=4 z 3x=6 x y z=
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∵|2x-3y+1|+(x+3y+5)的二次方=0∴2x-3y+1=0x+3y+5=0x=-2y=-1∴(-2x*y)的二次方(-y的二次方)×6xy平方的值=4x⁴y*(-y²
5X-4Y+4Z=13①2X+7Y-3Z=19②3X+2Y-Z=18③将③*4+①得17X+4Y=85④将③*3-②得7X-Y=35⑤将⑤*4+④得X=5将X=5代入①得Y=0将X=5,Y=0代入③得
x3y+2x2y2+xy3=xy(x2+2xy+y2)=xy(x+y)2,∵x+y=5,∴(x+y)2=25,x2+y2+2xy=25,∵x2+y2=13,∴xy=6,∴xy(x+y)2=6×25=1
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)=(
原式=4x29y2•27y364x3•4xy=34x2.故答案为34x2.
反应前XY均为0价,反应后化合价有变化,四氧化还原反应.提一句,4X2+Y2=X3Y+Y2去掉Y2的话是4X2=X3Y,这是不可能的,元素本身发生了变化,应该是核反应
∵x+y=4,∴(x+y)2=16,∴x2+y2+2xy=16,而x2+y2=14,∴xy=1,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=14-2=12.
∵3x=4y=5z,∴x20=y15=z12,设x20=y15=z12=k,∴x=20k,y=15k,z=12k,∴2x−3y+6z3x−2y+4z=2×20k−3×15k+6×12k3×20k−2×
(2x4-4x3y-x2y2)-2(x4-2x3y-y3)+x2y2=2x4-4x3y-x2y2-2x4+4x3y+2y3+x2y2=2y3,因为化简的结果中不含x,所以原式的值与x值无关.
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
已知x+y=5,xy=3,代数式x3y-2x平方y平方+xy3=xy(x²-2xy+y²)=xy(x-y)²=3×[(x+y)²-4xy]=3×(25-12)=
联立3x−4y−z=0①2x+y−z=0②,①-②得,x=5y③,把③代入②得,10y+y-z=0,解得z=11y,∴x+2y−3z3x−2y+3z=5y+2y−3•11y3•5y−2y+3•11y=
∵|x+y+1|≥0,|xy-3|≥0|x+y+1|+|xy-3|=0,∴x+y+1=0,即x+y=-1xy=3xy3+x3y=xy(x²+y²)=yx[(x+y)²-2
应该是X3y-2x2y2+xy3原式=x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=17/36*6=17麻烦采纳,谢谢!
x+y=4,xy=2后者平方后二式相加再加后者平方
x3y+xy3=xy(x^2+y^2)=(√3-√2)(√3+√2)((√3-√2)^2)+(√3-√2)^2)=1*(3-2√6+2+3+2√6+2)=10
∵x+y=3,∴(x+y)2=9,即x2+y2+2xy=9①,又x2+y2-3xy=4②,①-②,得5xy=5,xy=1.∴x2+y2=4+3xy=7.∴x3y+xy3=xy(x2+y2)=7.故答案
1.x/2=y/4=z/5可以推出y=2xz=5x/23x+4y+2z=163x+4×2x+2×5x/2=1616x=16x=1y=2z=5/22.2x+2y-4z=0①2x+4y+5z=0②①+②得
∵x-y=l,xy=2,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=2×1=2.
a1=a2,b1=-b2,c1=c2a1=a2,b1=b2,c1不等于c2a1不等于a2,b1不等于b2,c1不等于c2