已知2x-5y=0求2x y x的二次方-2xy y-搜答案-智慧作业帮

计算：（1）2xyx

（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

5x-4y=0;x=4y/5;x^2+y^2/x^2-y^2-x+y/x-y=（x²+y²-(x+y)²）/(x²-y²)=(x²+y

已知xyx+y＝2

∵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

x+y=02x+1=0x=-0.5y=0.52(x+5y)=2*(-0.5+2.5)=4平方根=±2

2x-3y=0(1)3x-y-1=0(2)由（2）得y=3x-1(3)把y=3x-1代入(1)得2x-9x+3=0∴x=3/7把x=3/7代入（3）得y=2/7∴x=3/7y=2/7

∵|x-3|+（y-x-5）2=0，∴x−3＝0y−x−5＝0，解得x＝3y＝8．

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

∵x²+y²-4x+2y+5=0x²-4x+4+y²+2y+1=0(x-2)²+(y+1)²=0x-2=0y+1=0∴x=2y=-14x&#

∵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

3x-2y5=x+y3交叉相乘,乘积相等：5(x+y)=3(3x-2y)5x+5y=9x-6y11y=4x4x=11y那么x：y=11：4

x-3y=0x=3y2x+5y/3x-2y=11y/7y=11/7

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

√x方-y+√2x+y=0x²-y=0,2x+y=02x+x²=0x(x+2)=0x=0或x=-2x=0y=0x+y=0x=-2y=x²=4x+y=-2+4=2

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