y=x-1分之x的平方减一的绝对值与y=kx-2有两个焦点

[（x的平方+y的平方）-（x-y)的平方+2y(x-y)]/4y=1(x²+y²-x²+2xy-y²+2xy-2y²)/4y=1(4xy-2y

y=(x^2-1)/(x^2+1)y(x^2+1)=x^2-1(y-1)x^2=-y-1x^2=(-y-1)/(y-1),由x^2>=0得(-y-1)/(y-1)>=0,所以(y+1)/(y-1)

X^2+Y^2=1表示圆,(Y-2)/(X-1)表示过点（1,2）的直线的斜率所以这道题是求过点（1.2）与圆相切的的直线的斜率.设直线为y-2=k(x-1)y=kx-k+2带入圆,x^2+(kx-k

根号下大于等于0x²-1>=0,x²>=11-x²>=0,x²

因为x=1+根号2y=1减根号2所以xy=1+根号2乘以1减根号2=1-2=-1x+y=2所以可以求得,原式=xy/(x+y)=-1/2

2x分之1减x+y分之1*（x的平方减y的平方+2x分之x+y）=2x分之1减x+y分之1*【（x+y）*（x-y）+2x分之x+y】=2x分之1-【x+y分之1*（x+y）*（x-y）+x+y分之1

(x分之x的平方-y的平方)乘以(x的平方-2xy+y的平方分之2x)其中x=2y=1=(x^2-y^2)/x*2x/(x-y)^2=(x+y)(x-y)/x*2x/(x-y)^2=2(x+y)/(x

x的平方-y的平方分之x的平方+2xy+y的平方除一x-y分之x的平方+xy,其中x=√2-1,y=√2+1=(x+y)^2/(x+y)(x-y)/((x^2+xy)/(x-y))=(x+y)/(x-

x/y=1/2y=2x(x^2+2xy+y^2)/(x^2-xy+y^2)=(x^2+2x*2x+(2x)^2)/(x^2-x*2x+(2x)^2)=9x^2/3x^2=3

y分之x=2分之1y=2x所以原原式=(x²-2x²-8x²)分之（x²＋4x²＋4x²）=-9分之9=-1

3∵x/y=1/2∴y=2x故（x^2+2xy+y^2)/(x^2-xy+y^2)=(x+y)^2/(x^2-xy+y^2)=(x+2x)^2/(x^2-x·2x+4x^2)=9x^2/(3x^2)=

（x/2y）²×（y/2x）-x/y²÷（2y²/x）=x²/2y×（y/2x）-（x/y²）×（x/2y²）=x/8y-x²/（

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

x的平方+2xy+y的平方分之x的平方-y的平方=(x+y)(x-y)/(x+y)²=(x-y)/(x+y)∵x+y=2,x-y=1∴原式=1/2

（1-y+x分之y）÷y的平方-x的平方分之x=[(y+x-y)/(y+x)]÷x/(y²-x²)=x/(y+x)*[(y+x)(y-x)/x]=y-x

x的平方-64y的平方分之2x-x-8y分之1=[2x-(x+8y)]/(x-8y)(x+8y)=(x-8y)/(x-8y)(x+8y)=1/(x+8y)

y分之x=2分之1y=2x所以原原式=(x²-2x²-8x²)分之（x²＋4x²＋4x²）=-9分之9=-1

原式=(X-Y)²/(X+Y)(X-Y)=(X-Y)/(X+Y)=(√3+1-√3+1)/(√3+1+√3-1)=2/2√3=1/√3=√3/3

x-y分之2x的平方-y-x分之x的平方-4xy+x-y分之2y的平方-x的平方=2x/(x-y)-(x²-4xy)/(y-x)+(2y-x²)/(x-y)=(2x+x²

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