arctan(x y) (1-xy)

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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

(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)^

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?

z=x*arctan(xy),求(dz/dx)|(1,1),(dz/dy)|(1,1)

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

求z=1+根号下ln(xy)的偏导数,求求u=arctan(m的平方乘n)的偏导数.

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

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

设z=arctan(xy),y=e的x次方,求dz/dx

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

(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

证明arctanx+arctany=arctan(x+y/1-xy),其中xy不等於1

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

用函数的近似公式证明当|x|与|y|都很小时,证明arctan{(x+y)/(1+xy)}≈x+y我想知道构造一个什么样

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

(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,