y=arctan3^2÷x-1的复合

tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(arctan1/5+arctan3)=（1/5+3）/（1-3/5）=8

证明:设arctan1+arctan2+arctan3=x那么tanx=tan(arctan1+arctan2+arctan3)=(tan(arctan1+arctan2)+tan(arctan3))

设arctan1+arctan2+arctan3=x那么tanx=tan(arctan1+arctan2+arctan3)=(tan(arctan1+arctan2)+tan(arctan3))/(1

设tanA=1.tanB=2,tanC=3,D=A+BtanD=tan（A+B）=(1+2)/(1-1*2)=-3tan(A+B+C)=tan(D+C)=(-3+3)/(1+9)=0=>A+B+C=1

因为arctan1=π/4只要证明arctan2+arctan3=3π/4即可,因为tan（arctan2+arctan3）=（2+3）/（1-2*3）=-1又π/4

两边同时取正切tan(arctan2/x+arctan3/x)=tan(45°)(tan(arctan2/x)+tan(arctan3/x))/(1-(tan(arctan2/x)*tan(arcta

[(x+2y)^2-(x+y)(3x-y)-5y^2]/2x=(x^2+4xy+4y^2-3x^2-3xy+xy+y^2-5y^2)/2x=(-2x^2+2xy)/2x=-x+y=2+1/2=二分之五

[(x+2y)²-2(x+y)(x-y)-5y²]÷(2x)=x²+4xy+4y²-x²+y²-5y²=4xy当x=-2,y=1/

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

原式=(x-y)²/(x+y)²(x+y)^4/(x-y)²*1/(x+y)=x+y=-3

先一个一个的展开括号项,再同项合并就行了啊[x(x-y)-y(x-y)+(x+y)(x-y)]÷2x=[xx-xy-(xy-yy)+x(x-y)+y(x-y)]÷2x=[xx-xy-xy+yy+xx-

我用x2表示的x的平方1,(x-y)2=x2-2xy+y2,(x+y)(x-y)=x2-y2,所以原式=(2x2-2xy)/2x=x-y=12,=xy+y2+x2-y2-x2=xy=-13,=x2-x

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

/>（1/x-y+1/x+y）÷x^2y^2/x^2-y^2＝（2x)/(x^2-y^2)÷(x^2y^2)/(x^2-y^2)＝（2x)/(x^2-y^2)÷(x^2-y^2)/(x^2y^2)＝2

再答：再答：看后面这个图再答：不懂可以问我

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

设a=arctan3,则tana=3b=arctan2,则tanb=2tan(arctan3-arctan2)=tan(a-b)=(tana-tanb)/(1+tanatanb)=1/(1+6)=1/

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

{(x²+y²)-(x-y)²+2y(x-y)}÷4y=1{x²+y²-(x²-2xy+y²)+2xy-2y²}÷4y=