大师作文网limln(1 xy) xy^(1 2)
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1/2≥(2/xy)^2==》√2/2≥|2/xy|==》√2≤|xy/2|==》2√2≤|xy|==》xy≤-2√2或2√2≤xy
1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
3xy[2xy-x(y-2)+x-1]=3xy(2xy-xy+2x+x-1)=3xy(xy+3x-1)=3x^2y^2+9x^2y-3xy
xy²+2xy-x=x(y²+2y-1)=x(y²+2y+1-2)=x[(y+1)²-2]=x(y+1+√2)(y+1-√2)再问：太厉害啊，那再做多两题吧，也
xy-1+x-y=XY+X-Y-1(加法交换律）=（XY+X）－（Y＋1）＝X（Y＋1）－(Y+1)（提取公因式X）=(Y+1)(X-1)（提取公因式Y+1）这样可以么?
5x^2y/-1/2xy*3xy^2=-7.5x²y²
3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2
该题为隐函数求导.xy+e^(xy)=1则y+xy'+e^(xy)(y+xy')=0解得：y'=-y/x解答完毕.
（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
因式分解xy²-2xy+x+2y-1解xy²-2xy-y²+x+2y-1=xy²-2xy+x-y²+2y-1=x(y²-2y+1)-(y&#
原式=1-3xy+x+(-4x-6xy)=1-3xy+x-4x-6xy=1-3x-9xy
根据平方差公式得,X^2Y^2-1.请采纳,谢谢
设t=xy+1;(xy+1)(x+1)(y+1)+xy=t(x+1)(y+1)+xy=t(xy+x+y+1)+xy=t(t+x+y)+xy=t^2+xt+yt+xy=(t+x)(t+y)-xy+xy=
(xy+1)(x+1)(y+1)+xy展开(x+1)(y+1)展开,得(xy+1)(xy+x+y+1)+xy即(xy+1)(xy+1+x+y)+xy将(xy+1)当做一个整体,展开得(xy+1)^2+
(xy+1)(x+1)(y+1)+xy=(xy+1)(xy+1+x+y)+xy=(xy+1)(xy+x+1)+y(xy+1)+xy=(xy+1)(xy+x+1)+y(xy+1+x)=(xy+x+1)(
(xy+1)(x+1)(y+1)+xy展开(x+1)(y+1)展开,得(xy+1)(xy+x+y+1)+xy即(xy+1)(xy+1+x+y)+xy将(xy+1)当做一个整体,展开得(xy+1)^2+
=(x²y²-4-x²y²-xy+4)/(-2xy)=-xy/(-2xy)=1/2
通分原式=[(yz+xz+xy)/xyz]×(xy)/(xy+yz+zx)=xy(yz+xz+xy)/[xyz(xy+yz+zx)]=1/z
平方差公式（xy+1）（xy-1）=x²y²-1如还有新的问题,请不要追问的形式发送,另外发问题并向我求助或在追问处发送问题链接地址,