已知x,y满足x分之1-y分之1=2

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

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

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

∵二分之x+1=四分之y+3∴2x+2=y+3即y=2x-1∴5分之x+2x-1=4分之2x+2解出来x=7,y=15把x,和y带进去一算就行了.

(x+1)/3=(y+2)/2=(x+y)/2x=2y=0

若要√（X-1）和√（1-X）都有意义,则必须x=1所以Y＜1/2故1-Y>0所以y-1分之1*根号（1-Y）平方=-1

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-

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

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

有理数x,y满足1/2|x-2y|=-2|x-4|左右乘以2得：|x-2y|=-4|x-4|移项得：|x-2y|+4|x-4|=0绝对值的和相加为0,则必定是0+0=0即|x-2y|=0,4|x-4|

x+y=1所以1/x+1/y=(1/x+1/y)(x+y)=1+y/x+x/y+1=(y/x+x/y)+2y/x>0,x/y>0所以y/x+x/y≥2√(y/x*x/y)=2所以1/x+1/y≥2+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

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

1/x-2/y=225/x+7/y=22一式*5-二式得-17/y=88y=-17/88x=17/198

条件应该是xyz=8吧,假设是,如下：1/x+1/y+1/z=(yz+xz+xy)/xyz=(yz+xz+xy)/8将z=-x-y代入上式,也可代入x=-z-y或y=-z-x得分子:y(-x-y)+x

等于0?(x+2y)^2-x-2y+1/4=0(x+2y-1/2)^2=0x+2y-1/2=0x+2y=1/2

1/x+1/y=-1/(x+y)两边同时乘x+y得1+y/x+1+x/y=-1所以y/x+x/y=-3