已知4y^ x^-4xy x-2的绝对值

（1）原式=2xy+x(x−y)+y(x+y)x2−y2=(x+y)2(x+y)(x−y)=x+yx−y；（2）原式=2a−(a+2)(a+2)(a−2)a−2(a+2)(a−2)=1a+2；（3）原

∵xyx+y=2∴xy=2（x+y）∴原式=3x−5×2(x+y)+3y−x+3×2(x+y)−y=−7x−7y5x+5y=−75

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)=

2X-Y/X+Y=2X+Y/2X-Y=1/2原式＝[2(2X-Y)/X+Y]-[4(X+Y)/2X-Y]=2*2-4*(1/2)=4-2=2P.S.第一次为团队答题,好鸡东哦o(∩_∩)o有什么不懂的

(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

已知y/x=2,则求代数式x+y/4x-5y的值分子分母同除以xy；=(1+y/x)/(4-5y/x)=(1+2)/(4-5×2)=3/(-6)=-1/2;您好,很高兴为您解答,skyhunter00

4x-2y=0→y=2x,把所有的y都用2x替代.结果是-1/7

解；已知正数x，y满足，x2+y2=1，则1=x2+y2≥2xy，∴xy≤12…①   又xyx+y=11x+1y≤12 1x•1y=xy2…②①②联立得xyx

3x-4y/2x+y=1/26x-8y=2x+y4x=9yx/y=9/4

答案是2.4x-2y/x+3y=(2x-y/x+3y)乘以2=4,4x+12y/2x-y=(x+3y/2x-y)乘以4=1/2x4=2,所以代数式等于4-2=2.

∵xyx+y=-2，yzy+z=43，zxz+x=-43，∴1x+1y=-12，1y+1z=34，1z+1x=-34，∴2（1x+1y+1z）=-12，即1x+1y+1z=-14，则xyzxy+yz+

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

证明函数f(x,y)=(x+y)/(x-y)在点(0,0)处的二重极限不存在.当点(x,y)沿着直线y=kx(k为不等于1的任意实数)趋于(0,0)时,limf(x,y)=lim(x+kx)/(x-k

把分式xyx+y中的x和y都扩大2倍后得：2x•2y2(x+y)=4xy2(x+y)=2•xyx+y，即分式的值扩大2倍．故选：B．

x=±1,y=±3,z=±2xyzz>y则0>x>z>yx=-1,y=-3,z=-2,x2y-[4x2y-(xyz-x2z)-3x2z]-2xyx=x2y-4x2y+xyz-x2z+3x2z-2xyx

(x-y)/(x+y)=4所以(x+y)/(x-y)=1/4(x-y)/[2(x+y)]-(4x+4y)/(x-y)=(1/2)*[(x-y)/(x+y)]-4[(x+y)/(x-y)]=(1/2)*

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=

x的平方+y的平方-2x+4y=-5x²－2x＋1＋y²＋4y＋4=0﹙x－1﹚²＋﹙y＋2﹚²＝0x-1=0,y+2=0x=1,y=-2

y=4x/3所以(2x-y)/(x+y)=(2x-4x/3)/(x+4x/3)=(2x/3)/(7x/3)=2/7

已知x²+y²+5/4=2x+y则x²-2x+1+y²-y+1/4=0即(x-1)²+(y-1/2)²=0x=1y=1/2所以y-x=1/2