2.x^1 2 y^1 2=a^1 2,设x=acos^4θ,θ为参数
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把x=-2y代入5x+2y=12得-8y=12y=-3/2所以x=3x+y-a=03+(-3/2)-a=0a=3/2a等于3/2时x=-2y
(x+y)(x-y)=12因为xy为自然数所以x=4y=2
{x-y=2①{2x+y=4②①+②得3x=6x=2把x=2代入①得y=0{x-y=1①{3x-y=5②②-①得2x=4x=2把x=2代入①得y=1{x-y=8①{3x+y=12②①+②得4x=20x
令a=x²+y²则a²-a=12a²-a-12=0(a-4)(a+3)=0a=4,a=-3平方数大于等于0所以a=-3不成立所以x²+y²=
x-4y=0,x²y²2x-12=0相交于A,B两点则直线AB
x(x-y)-y(y-x)=12那么:化简得:(x+y)*(x—y)=12x、y是自然数,所以x1=2,y1=4x2=4,y2=2所以x+y-xy=-2
12-2(2x^y-3x+4y)+3(x^2y+4x-3y)=12-4x²y+6x-8y+3x²y+12x-9y=-x²y+18x-17y+12当x=1,y=-2时原式=
已知(7x-y-a)²+|x-12|=0则7x-y-a=0,x-12=0x=1284-y-a=084-a=y>0a再问:再问:这道会吗?再问:儿子问我,自己也有点忘了再答:(1)a+2=0a
把x=-2y代入5x+2y=12得-8y=12y=-3/2所以x=3x+y-a=03+(-3/2)-a=0a=3/2a等于3/2时x=-2y
∵B={y|y=12x2−x+52,0≤x≤3},∴y=12x2-x+52=12(x-1)2+2,∵0≤x≤3,∴2≤y≤4,即B=[2,4]∵A={y|y2-(a2+a+1)y+a(a2+1)>0}
x=7y=5再问:7和5是怎样求出来的
9(x+y)^(2m)*(x-y)^(4n)*[-(x+y)^2]=-9(x+y)^(2m+2)*(x-y)^(4n)∴a=92m+2=104n=12-n∴m=4n=12/5
x+y-(2x-y)(x+y)=(x+y)(1-2x+y)12(a^2-b^2)^2-3(a^2+b^2)(a+b)^2=12(a+b)²(a-b)²-3(a²+b&su
f(12)=f(3+9)=f(3)+f(9)=f(3)+f(3)+f(3)+f(3)=4a
1.x>=0时2x+y=10y>=0时由2得到x=12,带入1得到y=-14,和假设矛盾y
-12/x=-3x-12=-3x^2x^2=4x1=2x2=-2y1=-6y2=6∴A1(2,-6)A2(-2,6)为所求.
-3a^2(x-2y)^3+12a(2y-x)^4=-3a^2(x-2y)^3+12a(x-2y)^4=3a(x-2y)^3{4(x-2y)-3a}=3a(x-2y)(4x-8y-3a)
若x≤0,|x|=-x|x|+x+y=10y=10代入x+|y|-y=12得x=12>0矛盾,∴x>02x+y=10①若y≥0,x+|y|-y=x=12y=10-2x∴yx-2y=12②联立①②解得x
(1)∵集合A={y|y=log2x,x>1}={y|y>0},B={y|y=(12)x,x>1}={y|0<y<12},∴A∩B={y|0<y<12}.(2)由(1)知:CRA={y|y≤0},B=