已知x,y满足y分之x=5,求分式4x

2x-3y-1=0x-2y+2=0∴x=8y=5(3+y²)/(2x-5)是不是3+y²还是3y²=(3+25)/11=28/11

由条件可得：①2x+y=2②5x+2y=4①×2,得4x+2y=4③②-③,得x=0⑤将⑤代入①,得y=2∴所求代数式=(0+2×2+1)/(2×0-3×2+7)=5/1=5

2分之2x+y=12x+y=24分之5x+2y=15x+2y=4x=0y=22x-3y+7分之3x+2y+3=(3x+2y+3)/(2x-3y+7)=(2x2+3)/(-3*2+7)=7/1=7

∵2x(y-1)+4(1/2x-1)=0∴xy=2∵(x+2y)(x-2y)=-5(y^2－6/5)∴x²+y²=6所以：（1）(x-y)²=x²+y²

实数x,y满足√2x-3y-1+|x-2y+2|=0因为√2x-3y-1与|x-2y+2|都≥0所以,2x-3y-1=0x-2y+2=0y=5,x=84y/(4x-5)=20/(32-5)=20/27

|X-Y|+|Y-5|=0推X-Y=0;Y-5=0Y=5.X=5

2x-3y-1=0x-2y+2=02x-4y+4=0y-5=0y=55y=25x=82x-4=125y/(2x-4)=25/12

已知x,y满足方程组x方+2y=5,(1)x-y=-1(2)(1)+(2)*2x²+2x-3=0解得x1=1x2=-3代入(2)y1=2y2=-21.x1=1y1=2时x+y分之x的三次方-

2y+3除以-3y+7这就是最后结果了.

已知x,y满足x^2+y^2+5/4=2x+y,求代数式xy/(x+y)x^2+y^2+5/4=2x+yx^2-2x+1+y^2-y+1/4=0(x-1)^2+(y-1/2)^2=0x-1=0,x=1

x方+y方+5/4=2x+y(x方-2x+1)+（y方-y+1/4)=0(x-1)方+（y-1/2)方=0x=1,y=1/2xy/(x+y)=1/3

解；由题意得2x-3y-1=0x-2y+2=0解之得x=8y=5所以2x-5分之3y的平方根=根号13

x^2+y^2-4x-2y+5=0,即(x-2)^2+(y-1)^2=0所以x=2,y=1所以√(x+y)/√(8y-2x)=√3/2望采纳

已知x,y满足√（2x-3y-1）+√（x-2y+2）=0,则√（2x-3y-1）=0,且√（x-2y+2）=0∴{2x-3y-1=0x-2y+2=0解得{x=8y=5∴2x-4分之5y=12分之25

2分之3x+y=14分之5x-y=1,3x+y=2(1)5x-y=14(2)由(1)+(2)式得8x=16x=2y=-42x-3y+7=2*2-3*(-7)+7=32

x平方+y平方-4x-2y+5=0(x-2)^2+(y-1)^2=0(x-2)^2=0(y-1)^2=0x=2y=1根号8y-2x分之（根号x）+y=(根号2)+1/根号(8*1-2*2)=[(根号2

x^2+y^2+5/4=2x+y(x-1)^2+(y-1/2)^2=0x=1,y=1/2xy/(x+y)=1*(1/2)/(1+1/2)=1/3

已知xy满足x²+y²+5/4(四分之五)=2x+y,求代数式(x+y)xyx²+y²+5/4=2x+yx²-2x+1+y²-y+1/4=0

将x^2+y^2+5/4=2x+y化为将x^2-2x+1+y^2-y+1/4=0即(x-1)^2+(y-1/2)^2=0上面的等式成立,则有x-1=0且y-1/2=0即x=1,y=1/2所以xy/(x

X-2y+z=0转换形式,得X=2y-Z,把它带入7X+4y-5z=0中,得7（2y-z）+4y-5z=0得18y-12z=0即y=2/3z把y=2/3z代入X=2y-Z中,得X=1/3z把X=1/3