若x y等于1,且(x 2)(y 2)等于3,求x的二次方
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(x2+y2)/(x-y)=(x2+y2-2xy+2xy)/(x-y)因为xy=1,所以=[(x-y)^2+2]/(x-y)=(x-y)+2/(x-y)因为x>y>0所以(x-y)>0所以有不等式的定
设x2-xy+y2=Ax2-xy+y2=A与x2+xy+y2=1相加可以得到:2(x2+y2)=1+A(1)x2-xy+y2=A与x2+xy+y2=1相减得到:2xy=1-A(2)(1)+(2)×2得
∵x²-2xy-3y²=0∴*(x-3y)(x+y)=0∴x=3y或x=-y(1)2x+y/x-2y当x=3y时,原式=7y/y=7当x=-y时,原式=y/(-3y)=-1/3(2
x=p*cos(d)y=p*sin(d)1
①x2+2xy-y2不符合完全平方公式的特点,不能用完全平方公式进行因式分解;②-x2-y2+2xy符合完全平方公式的特点,能用完全平方公式进行因式分解;③x2+xy+y2不符合完全平方公式的特点,不
设x^2-xy+y^2=Ax^2-xy+y^2=A与x2+xy+y2=3相加可以得到2(x^2+y^2)=3+A(1)x^2-xy+y^2=A与x2+xy+y2=3相减得到2xy=3-A(2)(1)+
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.
∵(x+y+z)2=x2+y2+z2+2xy+2yz+2xz,∴m=12[(x+y+z)2-(x2+y2+z2)]=12[(x+y+z)2-1]≥-12,即m有最小值,而x2+y2≥2xy,y2+z2
由题意(1-xy)(1+xy)=1-x2y2,∴只要求出x2y2的范围即可,∵x2+y2=1≥2x2y2,∴x2y2≤14,-x2y2≥-14,∴(1-xy)(1+xy)=1-x2y2≥1-14=34
∵|x|=3,∴x=±3;∵y2=16,∴y=±4;∵|x+y|=1,∴x+y=±1;∴x=3,y=-4或x=-3,y=4.∴xy=-12;∴A=4x2-2xy+4y2=4(x+y)2-10xy=4+
X2-1-2xy+y2=X2-2XY+Y2-1=(X2-2XY+Y2)-1=(X-Y)2-1=(X-Y-1)(X-Y+1)(x2+4y2)2-16x2y2=(x2+4y2)2-(4XY)2=(X2+4
∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6
设x2-xy+y2=M①,x2+xy+y2=3②,由①、②可得:xy=3−M2,x+y=±9−M2,所以x、y是方程t2±9−M2t+3−M2=0的两个实数根,因此△≥0,且9−M2≥0,即(±9−M
左边=[(x-y)^2+2xy]/(x-y)=(x-y)+2/(x-y)>=2根号二------均值不等式,其中x-y>0
6x2-xy-15y2=(2x+3y)(3x-5y)=0,所以x=-3/2y或x=5/3y
=[x+(3-√5)/2*y][x-(3-√5)/2*y]有点牵强,但这是唯一的答案了
x2+xy+y2=1,∴xy=1-(x2+y2),又−x2+y22≤-|xy|≤xy≤|xy|≤x2+y22,知−x2+y22≤1-(x2+y2)≤x2+y22,得出23≤x2+y2≤2.故选D
∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a
(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
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²