求y=arctanx 1分之x-1的导数

y+z分之x=x+z分之y=x+y分之z;合比定理：x/(x+y+z）=y/(x+y+z)=z/(x+y+z);得x=y=z,y+z分之x的值=1/2

(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2

2x-y分之x+y=2,则2x-y=2(x+y)4x-2y分之x+y-4x+4y分之2x-y=2(2x-y)分之x+y-4(x+y)分之2x-y=4(x+y)分之x+y-4(x+y)分之2(x+y)=

1/x-1/y=2001则(x-xy-y)/(x-y)=2002/2001

将分式左右交叉相乘得3(x+y)=5(3x-2y)3x+3y=15x-10y12x=13yx:y=13:12

x+y分之4x-2y-2x-y分之4x+4y=2(2x-y)/(x+y)-4(x+y)/(2x-y)=2*2-4*(1/2)=4-2=2

设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则（1）x+y-z=kz（2）x-y+z=ky（3）-x+y+z=kx（1）+（2）+（3）得x+y+z=k(x+y+z)∴k=1时,

2/3

解原式=2(x-y)/(x+y)-(x+y)/3(x-y)=6(x-y)(x-y)/[3(x+y)(x-y)]-(x+y)x+y)/[3(x-y)(x+y)]=5(x-y)(x-y)/[(x+y)(x

因为X+2/X-Y=2所以X=2Y+2将X=2Y+2代入X+Y/3（X-Y）-（X-Y）/2（X+Y）=（3Y+2）/3（2Y+2-Y）-（2Y+2-Y）/2（3Y+2）接下来自己算

tana=x1,tanb=x2tan(a+b)=(tana+tanb)/(1-tanatanb)=(x1+x2)/(1-x1x2)=-3√3/(1-4)=√3因为x1+x2=-3√3,x1x2=4所以

设a=arctanx1b=arctanx2则要求值为a+btana=x1tanb=x2tan(a+b)=(tana+tanb)/(1-tana*tanb)根据韦达定理x1+x2=-3√3/2x1*x2

令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0

x+y分之x-y=2求x+y分之3x-3y减x-y分之2(x+y)的值x+y分之x-y=2x-y=2(x+y)(x+y)分之(3x-3y)-(x-y)分之2(x+y)的值x+y分之3x-3y=(x+y

x+y分之x-y=2x-y=2(x+y)求x+y分之3x-3y=(x+y)分之3(x-y)=(x+y)分之6(x-y)=6-x-y分之2(x+y)=-(x+y)分之2(x+y)=-2

y=4x带入3x-y=3x-4x=-xx+y=x+4x=5x(x+y)/(3x-y)=5x/(-x)=-5

原式等价于10(x+1)=5(y+3)=4(x+y)10(x+1)=5(y+3)可得2x-y=15(y+3)=4(x+y)可得4x-y=15解得x=7y=13x+2y+3=363x+2y+1=48两式

5x=4y另x=4t,y=5t所以x/y=4t/5t=4/5(x+y)/y=(4t+5t)/5t=9/5再问：为什么5X=4Y再答：因为5x-4y=0移项5x=4y再问：为什么要设4T和5T再答：为了

解:不防设x=2A,则y=3A,z=5A.由x+y+z=20,可知2A+3A+5A=20,10A=20,A=2.则x=4,y=6,z=10.

y’=[sinx/(1-x²)]'=[(sinx)'(1-x²)-sinx(1-x²)']/(1-x²)²=[cosx(1-x²)+2xsi