若x,y满足(x-1)^2+(y+2)^=4,求S=2x+y的最大值和最小值
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/14 21:43:44
xPN0|4/U`R-EPU(CVaE?AM}gulNQ7ieP11f?SJ,c+KK=ֽN 蠘 ,\a*C;LCqBCRVfIo:RoTGkvN視,
z@!B Al0ڬg%dџAEpwI.П5SO\,"2|$Y
S:Gw8
若变量x,y满足约束条件{x>=-1;y>=x;3x+2y
已知实数x,y满足y=|x-1|若x+2y
设x,y满足约束条件x+y>=1,x-y>=-1,2x-y
设x,y满足约束条件x+y>=1,x-y>=-1,2x-y
变量x,y满足约束条件,x+y>=3,x-y>=-1,2x-y
方程组2x+y=1-m...x+2y=2中,若未知数x,y满足x+y
若x,y满足|x-y+1|+(x+y+3)^2=0则x^2-y^2=
若x,y满足|2x-y-3|+|3x+2y+1|=0,则x= y=
若x、y满足2x+y=6,x-3y=1,求7y(x-3y)²-2(3y-x)³的值
若x.y满足x+=4和x-y=2,求xy分之2x-y+1除以-x分之y+y分之x的值
已知x,y满足约束条件:x-y+1>=0,x+y-2>=0,x
若x,y满足y
若关于x y的二元一次方程组2x+y=2m-1 x+2y=m满足x+y>0和x-y
若关于x,y的二元一次方程组x+2y=m-1,2x+y=m+1的解x,y满足x>0,y
若关于x,y的二元一次方程组x+2y=m-1,2x+y=m+1的解x,y满足x>0,y
若实数X,Y满足X²+Y²+4X-2Y +1=0,则X²-6X+Y²-2Y的最大值是多少
若x.y属于R,且满足(X*X+Y*Y+2)(X*X+Y*Y-1)-18≤0.求证:XY≤2
若实数x,y满足约束条件x-y≥0,x+y≥0,2x+y≤1,则y/(x+1)的最大值是多少,