若方k平方减1x平方加3y平方等于1,表示焦点在x轴上的椭圆,则实数k的取值范围-搜答案-智慧作业帮

x^2+2x+y^2-6y+10=0可以化解为（x+1)^2+(y-3)^2=0,等价于x+1=0且y-3=0,所以有x=-1,y=3,故x^3*y=-3.

x^2/n-y^2/(3-n)=1渐近线方程为x^2/n-y^2/(3-n)=0即(3-n)/n=4解得n=3/5

x/y=3x=3y(x^2+y^2-xy)/(x^2-y^2)=(9y^2+y^2-3y^2)/(9y^2-y^2)=7y^2/8y^2=7/8

∵(x+1)²+|y-2|=0　∴x+1=0　　y-2=0　∴x=-1　　y=2　∴x²y²+3xy-7x²y²-(5/2)xy+1+5x²

x²＋xy－2y²＝0﹙x－y﹚﹙x＋2y﹚＝0x＝y,x＝－2y当x＝y,﹙x²＋3xy＋y²﹚／﹙x²＋y²﹚=﹙y²＋3y&

x2+xy-2y2=0(x-y)(x+2y)=0x-y=0,x+2y=0x=y或x=-2yx=y,则原式=(y2+3y2+y2)/(y2+y2)=5y2/2y2=5/2x=-2y,则原式=(4y2-6

X2+Y2-4X-6Y+13=0X2-4X+4+Y2-6Y+9=0（X2-4X+4）+（Y2-6Y+9）=0（x-2）2 +（y-3）2=0所以,x-2=0 x=2y-

根据题意开口必向上而且判别式要小于0所以(k的平方加4k减5）大于0k小于-5或k大于1判别式=k^2-20k+19小于01小于k小于19综上k的取值范围是（1,19）

x²y-2x²y²+xy³=xy(x-2xy+y²)出题者估计是X的三次方y减2x的平方y的平方加xy的三次方x³y-2x²y&#

(y+1)平方-2x平方(1+y平方)+x四次方(1-y)平方=(y+1)²+2*x²*(y²-1)+[x²(y-1)]²-2x²(y

x+y=1则x^2+y^2+2xy=1因为x^2+y^2=3则xy=-1x^3+y^3=(x+y)(x^2-xy+y^2)=1*(3+1)=4

先分解因式,后乘法.(X^4-Y^4)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X+Y)(X^2+Y^2)/(X^3+X^2Y+XY^2+Y^3)=(X-Y)(X^3+X^2Y+XY^2+

xy=0.k=0.最后=-1

充要条件

6X的四次方减11

(x^2+y^2)^2-4x^2y^2=(x^2+y^2+2xy)(x^2+y^2-2xy)=(x+y)^2*(x-y)^2=(x^2-y^2)^2数学佰亊嗵为你解答,希望我的回答为您带来帮助!

x平方/(5-k)+y平方/（3-k）=1k-3>05-k>05-k>k-3得3

x²+y²=x²-xy+y²+xy=3+5=82x²+xy+3y²=2x²-2xy+3y²+3xy=2(x²-

x³-y³-x²y+xy²=(x³-y³)-(x²y-xy²)=(x-y)(x²+xy+y²)-xy

(x-3)²+|y-1|+z²=0平方数和绝对值永远大于等于0(x-3)²=0,|y-1|=0,z²=0x=3,y=1,z=0(x-y)²+(y-z)