若对圆(x-1)2 (y-1)2=1上任意一点
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![若对圆(x-1)2 (y-1)2=1上任意一点](/uploads/image/f/6974942-14-2.jpg?t=%E8%8B%A5%E5%AF%B9%E5%9C%86%28x-1%292+%28y-1%292%3D1%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9)
2−x≥a4−y,即a≤(2-x)(4-y)恒成立,只需a≤(2-x)(4-y)的最小值而(2-x)(4-y)=8-4x-2y+xy=8-(4x+2y)+2=10-(4x+2y)=10-(4x+4x)
B={(x,y)|(a^2-4)x+(a-2)y=16}当a=2,B为空集,符合题意当≠2时,)A集合中的元素是直线L1:(y-2)/(x-1)=a+2是的点除去点(1,2)B集合中的元素为直线L2:
(1)令x=-1,y=1,则由已知f(0)-f(1)=-1(-1+2+1)∴f(0)=-2(2)令y=0,则f(x)-f(0)=x(x+1)又∵f(0)=-2∴f(x)=x2+x-2(3)不等式f(x
(1)f(0)=-2(2)f(x)=x^2+x-2(3)由题意得,x^2+x-2+3-2x
x=1,y=0f(x+y)-f(y)=x(x+2y+1)f(1)-f(0)=2f(0)=-2(2)y=0f(x)-f(0)=x(x+1)f(x)=x^2+x-2
Dx^y+x^-y=2根号2===>(x^y+x^-y)^2=8===>x^2y+x^-2y+2=8===>x^2y+x^-2y=6(x^y-x^-y)^2=x^2y+x^-2y-2=6-2=4==>
不等式(x-a)⊙x≤a+2可化为(x-a)(1-x)≤a+2即-x²+x+ax-a≤a+2a(x-2)≤x²-x+2因为x>2所以 a≤(x²-x+2)/(x-2)令f
(1)∵f(x+y)-f(y)=(x+2y+1)x∴f(1+0)-f(0)=(1+2*0+1)*1即f(1)-f(0)=2∵f(1)=0∴f(0)=-2(2)∵f(x+y)-f(y)=(x+2y+1)
(1)令x=1y=0带入原式得f(0)=-2(2)令y=0带入原式得f(x+0)-f(0)=(x+2*0+1)*x所以f(x)=x平方+x-2将f(x)带入不等式得a>x平方-x+1当0
1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)(2a+b)(2
令x=y=0,带入f(0)+f(0)=0f(0)=0令x=y=1带入f(1)+f(1)=1f(1)=1/2令y=x带入f(x)+f(x)=x(2x-1)f(x)=x^2-x/2a<1f(x)=x^2-
1)函数y=f(x)对任意x,y属于R,恒有f(x+y)=f(x)+f(y)当X=0,Y=0时,有F(0)=F(0)+F(0)===>F(0)=0当Y=-X时候,有F(0)=F(X)+F(-X)==>
令x=y=0,则f(0)=2f(0),则f(0)=0,再令y=-x,则f(0)=f(x)+f(-x)=0,则f(x)=-f(-x),即y是奇函数.2.f(3)=-f(-3)=-a,f(24)=f(3)
y^4+2x^4+1=4x^2yy^4-2y^2+1+2(x^4-2x^2y+y^2)=0(y^2-1)+(x^2-y)=0y^2=1,x^2=yy1=1,y2=-1,(当y=-1时,x^2-y不能等
要使(x+2y)(2x+1y)≥m恒成立,只要使(x+2y)(2x+1y)的最小值≥m即可,∵(x+2y)(2x+1y)=2+2+4yx+xy≥4+2xy4yx=8∴8≥m故答案为(-∞,8]
(1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/
∵x>0,y>0∴x+y+3=xy≤(x+y2)2∴x+y≥6由(x+y)2-a(x+y)+1≥0可得a≤x+y+1x+y恒成立令x+y=t,f(t)=t+1t在[6,+∞)上单调递增,则当t=6时f
x+y+3=xy(x-1)(y-1)=4x+y≥6令x+y=tt²-at+1≥0恒成立1°a/20,y>0)的任意x,y(x+y)^2-a(x+y)+1≥0恒成立a的范围是a≤37/6
解题思路:同学你好。你给的题目,若x,y是整数,解答只能是这样。请再看看题目。我猜测:x,y是正数,解题过程:
设z=(x-1)/[(x-1)^2-y]=(x-1)/(x²-2x+1-y)则∂z/∂x=[1*(x²-2x+1-y)-(x-1)(2x-2)]/(x