x+y=x+x-(x-y+5)
已知O是坐标原点,x,y满足条件x-y+5≥0,x+y≥0,x≤3,且M(2,1),P(x,y).
1个回答
相关问题
-
已知点P(x,y)的坐标满足{x-y+1≥0,x+y-3≥0,x≤2},O为坐标原点,则|PO|的最小值为
-
已知O为坐标原点,点M(1,-2),点N(x,y)满足条件(x≥1,x-2y≤1,x-4y+3≥0),则向量OM与向量O
-
已知x、y满足条件7x-5y-23≤0,x+7y-11≤0,4x+y+10≥0,M(2,1),P(x,y).求:1.y+
-
若变量x,y满足约束条件(x-2y+1≤0,2x-y≥0,x≤1)则点P(2x-y,x+y)与坐标原点O所在直线斜率的范
-
已知M(x,y)满足条件『x>=0,y>=0,2x+y+k=0,y>=0,2x+y+k
-
已知点A(3,根号3),O为坐标原点,点P(x,y)满足:根号3x-y≤0,x-根号3y+2≥0,y≥0,求向量OA*向
-
已知圆x²+y²+x-6y+m=0和直线x+2y-3=0交于P、Q两点,且OP⊥OQ(O为坐标原点)
-
已知x,y满足方程组{-2x+5y=m,-2x+7y=3m,且m≠0,则x/y=
-
已知点P的坐标(x,y)满足x-4y+3≤0,3x+5y≤25,x-1≥0,及A(2,0),则|op|cos∠AOP(o
-
已知动点P(x,y)满足x^2+y^2-2|x|-2|y|=0,O为坐标原点,则|PO|的取值范围是