x^1/2+y^1/2=a^1/2,设x=acos^4θ,θ为参数.化成参数方程

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 设集合A={x|x=y^2-2y-8},B={x|y=-x^2+2x+3},C={y|y+1 设平面点集A={(x,y)|(y-x)(y-1/x)>=0},B={(x,y)|(x-1)^2+(y-1)^2 设平面点集A={(x,y)|(y-x)(y-1/x)>=0},B={(x,y)|(x-1)^2+(y-1)^2 设平面点集A={(x,y)|(y-x)(y-1/x)>=0},B={(x,y)|(x-1)^2+(y-1)^2 设集合A=｛y/y=x²-2x+3,x∈R｝,B=｛y/y=-x²+2x+10,x∈R｝,求A∩B设集合A=｛（x,y)/y=x+1,x∈R｝,B=｛（x,y)/y=-x²+2x+3/4,x∈R｝,求A∩B 设x大于1,y大于0,x^y+x^-y=2根号二,x^y-x^-y等于? 设y(x)=(1+x^3)(5-1/x^2),求y'(1),y'(a).把步骤写清,跪拜设y(x)=(1+x^3)(5-（1/x^2）),求y'(1),y'(a). 设集合A={x//x-2/≤2,x∈R},B={y/y=-x^2,-1 设y=xlnx/1+x^2, 设集合A={y丨y=1+4x-x^2,-1 设y=[a^(2x+1)]+3x,求y的导数 设y=arctan(a/x)+1/2[ln(x-a)-ln(x+a)],求dy|x=0 因式分解（x+y-2xy)(x+y-2)+(xy-1)^2设x+y=a,xy=b, 设集合A={(x,y)|y=x^2+mx+2},B={(x,y)|x-y+1=0,x 设集合A=｛x|y=log2(x-1)},B={y|y=-x^2+2x-2,x属于R}求A∩B 设集合A={x|y=根号(3x-x^2)} B={y|y=2^x,x>1}则A∩B