arctan(x y) (1-xy)-搜答案-大师作文网

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)^

步骤如下:

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?

dz/dx=arctan(xy)+xy/[1+(xy)^2](dz/dx)|(1,1)=π/4+1/2(dz/dy)|(1,1)=x^2/[1+(xy)^2]=1/2

1.az/ax=1/2*1/√ln(xy)*1/(xy)*y=1/(2x√ln(xy))同理：az/ay=1/(2y√ln(xy))2.au/am=1/(1+(m^2n)^2)*n*2m=2mn/(1

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

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

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

左右2边取正切,左边=(X+Y)/(1-XY)=右边.左边=arctan[(X+Y)/(1-XY)+Z]/[1-(X+Y)Z/(1-XY)]=arctanc(X+Y+Z-XYZ)/[1-XY-(X+Y

-xy+xy等于多少?

得0=(-X+X)y=0×y=0

当x.y趋近与零时tan(x+y)/(x+y)=1,1+xy=1.所以tan(x+y)/[(x+y)/(1+xy)]=1有些打不出来,这道题就是个极限题

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,