求y=2-(-x^ 4x)^0.5的值域.

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)=

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

x^2+y^2-4x-2y+1=0(x-2)^2+(y-1)^2=4x^2+y^2=4x+2y-1设(x-2)^2=k(y-1)^2=4-kx^2+y^2+x+y=5x+3y-1x=2+根号ky=1+

(2x-y)/(x+y)=2两边同时乘以22*(2x-y)/(x+y)=2*2把2乘以进去,有(4x-2y)/(x+y)=4(2x-y)/(x+y)=2两边取倒数,有(x+y)/(2x-y)=1/2两

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

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

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

x-(x+y)+(x+2Y)-(x+3y)+(x+4y)...-(x+2009y)=-y+2y-3y+4y+...-2007y+2008y-2009y=1004y-2009y=-1005y=-1005

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

已知x^2+4y^2-4x+4y+5=0求(（y^4-x^4）/(y-2x)(x+y))*((2x-y)/(xy-y^2))/((x^2+y^2)/y)的值答案：x²+4y²-4x

两边同时乘以分母,移项,看成是关于x的一元二次方程,yx^2+(2y-1)x+4y=0方程有解的条件是（2y-1)^2-4*y*4y大于等于0,解这个一元二次不等式得到.不好打了,自己求吧呵呵

4x+2乘过去整理得.x(4y-5)=-2y-1x=(-2y-1)/(4y-5)因为x

已知x²+y²+5=2x+4y所以(x-1)²+(y-2)²=0故x=1,y=2所以（2x²-（x+y)(x-y))×((x+y-1)(x-y+1)+

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

（1）x^2+y^2-4x+10y+29=0x^2-4x+4+y^2+10y+25=0(x-2)^2+(y+5)^2=0x=2y=-5剩下的自己解下面两个题目是不是写错了,没看懂

已知(2x-y)/(x+y)=2,则(4x-2y)/(x+y)=4,(x+y)/(2x-y)=1/2,(4x+4y)/(2x-y)=2(4x-2y)/(x+y)-(4x+4y)/(2x-y)=4-2=

把原方程整理得：2x^4－4x^2y＋y^4＋1＝0则这个关于x^2的一元二次方程有实数解,故得：(－4y)^2－8(y^4＋1)≥0即有：(y^2－1)^2≤0当且仅当y^2－1＝0时,上述方程有实

1,-3再问：过程。。。再答：★（x²－2x)+（y²－4y）=5★（x－1）²+（y－2）²=1+4－5★（x－l）²=0,（y－2）²=

对待这样的题,先画出条件里的二次曲线,然后用与y-x=0平行的直线与所给曲线相切,然后切点就是最值点

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009）y=0+(1×1004-2009)y=-1005y=1005/2