xy=4和y=4x和x=2图像的面积
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x^2+y^2=17xy=-4故x^2+y^2-2xy=(x-y)^2=17-2(-4)=25,且x>y,故x-y=5同理x+y=3故x^2-y^2=(x-y)(x+y)=3*5=15
x=3/2(6+9/2Y+2Y)/(-6+12Y-2Y)故这样的话这题是解不出来滴故原题我预测应该是1/X+2/Y=2有:Y+2X=2XY原式可以化为:(4XY+3XY)/(-4XY+8XY)=7/4
太费劲了.比较简单易懂的分析:x+xy+y=-1这个式子是个什么图形?原方程化为:(1+y)(1+x)=0,即x=-1和y=-1两条直线.下面分别求第一个方程在x=-1和y=-1时的交点.x=-1代入
已知x>2y,xy=1,故设x-2y=t>0,则(x-2y)²=t²→x²+4y²=t²+4xy=t²+4.∴依基本不等式得(x²
4x^2-xy-3y^2=0(4x+3y)(x-y)=04x+3y=0或x-y=0x=-3/4y或x=y当x=y时,(x+2y)/(x-2y)=3y/(-y)=-3当x=-3/4y时,(x+2y)/(
(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^
x和y是互为倒数,则xy=1x^2-5xy+3x+1=0x²-5+3x+1=0(x+4)(x-1)=0x=-4或x=1①x^2+xy-2y^2=0②x^2+3xy+4y^2=8①x^2-2y
2X^2-xy-5x+y+4=o可化为:(x^2-4x+4)+(X^2-XY-X+Y)=0(X-2)^2+(x-Y)(x-1)=0因为x>=YY>=1所以X-Y>=0x-1>=0又因为(X-2)^2>
∵2x2-xy-5x+y+4=0∴x2+x2-xy-4x-x+y+4=0∴x2-4x+4+x(x-y)-(x-y)=0∴(x-2)2+(x-y)(x-1)=0∵(x-2)2≥0,x≥y≥1,∴(x-y
(X-5)和(Y+2)²是互为相反数则|x-5|+(Y+2)²0则x-5=0,Y+2=0则x=5,Y=-23A-2B=3(2x²-3XY+Y²)-2(4X
x=-1,y=-6或x=6,y=1代入式子得x^2-xy^2=37或30x^2y+xy^2=-42或42
(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60
2X^2-XY-5X+Y+4=X^2+X(X-Y)-4X-X+Y+4=(X-2)^2+(X-1)(X-Y)=0x≥y≥1(x-2)^2>=0,x-1>=0,x-y>=0,即(x-1)(x-y)>=0两
你看一下 我做的有点急 不知道对不
实数x、y,满足x≥y≥1,2x^2-xy-5x+y+4=0,则2x^2-5x+4=(x-1)y≤(x-1)x,即2(x-2)^2≤0,x=2,y=2.
2x²-xy-5x+y+4=o(x²-4x+4)+(x²-xy-x+y)=0(x-2)²+(x-y)(x-1)=0因为x≥y≥1所以x-y≥0x-1≥0即(x-
后面那个是.前面的式子x-y+xy=1中的xy是二次的,所以是二元二次方程.
(4x+3y-1)2平方+|3-y|=0一个数的平方和一个数的绝对值都是非负数它们的和为0则它们分别为0则4x+3y-1=03-y=0则x=-2y=3则xy=-6x+y=1
1关于原点对称图像为双曲线在2、4象限2图像无限接近坐标轴但永不相交3图像关于原点对称4在每个象限里Y随X的增大而增大5第二象限的图像过点(-1,4)(-2,2)(-4,1)第四象限的图像过点(1,-
x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60