[xy 1][xy-1]

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[xy 1][xy-1]
1/2≥(2/xy)^2 求xy大小?

1/2≥(2/xy)^2==》√2/2≥|2/xy|==》√2≤|xy/2|==》2√2≤|xy|==》xy≤-2√2或2√2≤xy

先化简,再求值 ⒈2(Xy+Xy)-3(Xy-xy)-4Xy,其中X=1,y=-1

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

已知根号xy3与根号xy1互为相反数

解√x-y+3与√x+y-1互为相反数∴√x-y+3+√x+y-1=0∵√x-y+3≥0√x+y-1≥0∴x-y+3=0x+y-1=0两式相加2x+2=0∴x=-1,y=2∴(2x-y)=2×(-1)

(x+y-2xy)(x+y-2)+(1-xy)²

(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)^

计算3xy[2xy-x(y-2)+x-1]

3xy[2xy-x(y-2)+x-1]=3xy(2xy-xy+2x+x-1)=3xy(xy+3x-1)=3x^2y^2+9x^2y-3xy

1、xy²+2xy-x(因式分解)

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-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

5x^2y/-1/2xy*3xy^2=-7.5x²y²

当x=3,y=3分之1时,求代数出3xy-[2xy-2(xy-2分之3xy)+xy]+3xy的值

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

xy+e^(xy)=1,求y的导数

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.

已知:xy+x=-1,xy-y=-2.

(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

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&#

(xy+1)(xy-1)用平方差怎么算

根据平方差公式得,X^2Y^2-1.请采纳,谢谢

用换元法分解因式 (xy+1)(x+1)(y+1)+xy

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=

因式分解:(xy+1)(x+1)(y+1)+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怎么因式分解

(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)(

对(xy+1))(x+1)(y+1)+xy进行因式分解.

(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+

【(2+xy)(xy-2)-xy(xy+1)+4】/(-2xy) 一道计算题!

=(x²y²-4-x²y²-xy+4)/(-2xy)=-xy/(-2xy)=1/2

(1/x+1/y+1/z)×(xy)/(xy+yz+zx)

通分原式=[(yz+xz+xy)/xyz]×(xy)/(xy+yz+zx)=xy(yz+xz+xy)/[xyz(xy+yz+zx)]=1/z

(xy+1)(xy-1)的计算及其答案

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,