X2-y2=1,则12x2-4xy的最小值
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假设x^2+y^2=m那么m(m+1)=20即(m+5)(m-4)=0那么m=-5或4所以x^2+y^2=4
因为向量OC乘以向量OD等于0,oc与od垂直.即为(y1/x1)(y2/x2)=-1,y1和y2用x1和x2换掉就能代了.
由于1=x2+y2+z2=(x2+12y2)+(12y2+z2)≥2x•y2+2•y2•z=2(xy+yz),当且仅当x=y2=z时,等号成立,∴x=y2=z=12时,xy+yz的最大值为22.故答案
设x2+y2=t,则方程即可变形为t(t-1)-12=0,整理,得(t-4)(t+3)=0,解得t=4或t=-3(不合题意,舍去).即x2+y2=4.
1,16x2+9y2=144化为标准方程x²/(144/16)+y²/(144/9)=1x²/(12/4)²+y²/(12/3)²=1x&s
:(1)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=3x&
可以把x2+y2看成一个X(x>=0)即x(x-1)=12x2-x-12=0(x-4)(x+3)=0x1=4x2=-3即x2+y2=4x2+y2=-3(不合题意,舍去)
A、原式=-6x2-19xy-5y2;B、原式=2x2-9xy-7y2;C、原式=x2-16xy-10y2;D、原式=8x2-13xy-15y2.故选D.
∵(x2+y2+1)2-4=0,∴(x2+y2+1)2=4,∵x2+y2+1>0,∴x2+y2+1=2,∴x2+y2=1.故答案为:1.
原式=1-(x2-4xy+4y2)=1-(x-2y)2=(1+x-2y)(1-x+2y).故答案为(1+x-2y)(1-x+2y).
令x2+y2=t,原方程变形为,t(t-1)=2,整理得,(t-2)(t+1)=0,解得t1=2,t2=-1,∵x2+y2≥0,∴x2+y2=2.故答案为2.
(x²+y²)(x²+y²-1)-12=0(x²+y²)²-(x²+y²)-12=0[(x²+y&s
(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√
X2+xy-(xy+y2)=4-12x2+xy-xy-y2=-8x2-y2=-8x2+xy+xy+y2=4+12x2+2xy+y2=16
∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6
x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=
可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.
解题思路:椭圆解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.php?
x2+4x+y2-2y+5=0,x2+4x+4+y2-2y+1=0,(x+2)2+(y-1)2=0,x+2=0,y-1=0,解得x=-2,y=1,x2+y2=5,故答案为:5.
(1)∵x2-y2=x2+xy-xy-y2=x2+xy-(xy+y2)而x2+xy=2,xy+y2=-1,∴x2-y2=2-(-1)=3;(2)∵x2+2xy+y2=x2+xy+xy+y2,而x2+x