在方程(x 2y-8)
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在X2Y中,Y质量分数为40%,即Y%=Y/(2X+Y)=40%所以2X=3/2Y在YZ2中,Y质量分数为50%,即Y%=Y/(Y+2Z)=50%所以2Z=Y即Z=1/2Y所以3Z=3/2Y所以在化合
(1)4ab+8-2b2-9ab-6=-2b2-5ab+2(2)原式=3x2y-2x2y+6xy-3x2y+xy=-2x2y+7xy,当x=-1,y=-2时,原式=-2×(-1)2(-2)+7×(-1
设X、Y、Z的相对原子质量分别为a、b、c,由题可得b2a+b ×100%=40%,则a=0.75b,bb+2c ×100%=50%,则c=0.5b.在X2YZ3中Y的质量分数为:
5x2y+(-6x2y)+34x2y=14x2y答:和是-14x2y.
1,x:ay:bz:cb/(2a+b)=0.4b/(b+2c)=0.5X2YZ3b/(2a+b+3z)=0.25选C2.尿素CO(NH2)2选B.25*94.3%*(28/60)=11
∵代数式x3+y3+3x2y+axy2含有因式x-y,∴当x=y时,x3+y3+3x2y+axy2=0,∴令x=y,即x3+x3+3x3+ax3=0,则有5+a=0,解得a=-5.将a=-5代入x3+
Y/(2X+Y)=0.4;Y/(Y+2Z)=0.5,可得,3Y=4X,Y=2Z.所以YX2YZ3中的质量分数=2Y/(2X+2Y+3Z)=0.4
x2yy占40%=>x:y=3:4yZ2y占50%=>y:z=2:1===>x:y:z=3:4:2x2yz3中y占4/(3*2+4+2*3)=4/16=25%
设Y的相对原子质量为M,则根据X2Y中含40%Y可得X=(M/0.4-M)/2即X=0.75M,同理根据YZ3中含50%Y可得Z=(M/0.5-M)/3即Z=M/3,所以化合物X2YZ3中Y的质量分数
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^
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
答案:2x^2y+2xy^2原式=4x2y-{x2y-[3xy2-2x2y+4xy2+x2y]}-5xy2=4x2y-{x2y-[7xy2-x2y]}-5xy2=4x2y-{x2y-7xy+x2y]}
8x2y-8xy+2y,=2y(4x2-4x+1),=2y(2x-1)2.
5x2y+3x2y+(-4x2y)=(5+3-4)x2y=4x2y,故答案为:4x2y.
楼上兄的回答思路是正确的,只不过修正一下小错误symsxyf=sin(x^2*y)*exp(-x-y);ddf=diff(diff(f,x),y);simple(ddf)
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
x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)
x2y-2xy-y=y(x2-2x-1)=y(x2-2x+1-2)=y[(x-1)2-(2)2]=y(x-1+2)(x-1-2),故答案为:y(x-1+2)(x-1-2).
2x2y与3x2y是同类项.2x2y+3x2y=5x2y.