x^2 y^2=1,(4,2),y=2x-1截得的弦长为4

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 09:09:54
x^2 y^2=1,(4,2),y=2x-1截得的弦长为4
[(2x+y)^2+(y+2x)(y-2x)-2y(4x-y)]/4y,其中x=1/2,y=1/3

[(2x+y)^2+(y+2x)(y-2x)-2y(4x-y)]/4y=(4x^2+4xy+y^2+y^2-4x^2-8xy+2y^2)/4y=(-4xy+4y^2)/4y=-x+y=-1/2+1/3

化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

x,y属于R*,且x+y=1,求证:(1)(x+1/x)(y+1/y)≥25/4 (2)(x+1/x)^2+(y+1/y

第一题左边=xy+x/y+y/x+1/xy≥2+xy+1/xy其中xy∈[0,1/4]故由耐克函数的图像,知xy+1/xy≥17/4因为两次放缩可同时取到等号,故证明无误,即左边≥17/4+2=25/

若x=-1/4,能否确定代数式(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)的值?

(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)=4x^2-y^2+2xy-8x^2-y^2+4xy+2y^2-6xy=-4x^2=-4(-1/4)^2=-1/4

1、x(x-y)(x+y)-x(x+y)^2

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)(2a+b)(2

先化简,再求值(x-y/x+y)^2÷(y-x/x^2+2xy+y^2)^2*1/x+y,其中x=-4,y=1

原式=(x-y)²/(x+y)²(x+y)^4/(x-y)²*1/(x+y)=x+y=-3

3x-2y=1 4x-5y

条件不完全.补充后给你解答.4x-5y是4x=5y还是缺少了等号后的值?再问:4x-5y=3再答:

1) (2x+5y)(2x-5y)(-4x^2-25y^2) 2) [(x+y)(x-y)-(x-y)^2+2y(x-y

1)(2x+5y)(2x-5y)(-4x^2-25y^2)=-(4x^2-25y^2)(4x^2+25y^2)=-(16x^4-625y^4)=625y^4-16x^42)[(x+y)(x-y)-(x

化简求值:【(x+Y)(X-Y)-(X-Y)的平方+2Y(2-Y)】除4Y,其中X=1 Y=2

(x-y)[(x+y)+x-y]+4y-2y^2=2x(x-y)+4y-2y^2=2[x^2-xy+2y-y^2]所以=2[1-2+4-4]/8=-1/4

先化简再求值(x-y)(x+y)-(x-2y) 的完全平方+x(3x-5y)-(x-y)(x-2y),其中x=1/2 y

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

已知x²+y²+5=2x+4y,求代数式(2x²-(x+y)(x-y))x((x+y-1)

已知x²+y²+5=2x+4y所以(x-1)²+(y-2)²=0故x=1,y=2所以(2x²-(x+y)(x-y))×((x+y-1)(x-y+1)+

已知4x=9y求(1)x+y/y (2)y-x/2x

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

先化简,再求值:[(x+y)(x-y)-(x-y)^2+2y(x-3y)]/(-4y),其中x=1,y=-2

为你提供精确解答先化简:(x^2-y^2-x^2+2xy-y^2+2xy-6y^2)/(-4y)=(4xy-8y^2)/(-4y)=-x+2y=-1-4=-5其他的正在为你解答.

9y=3x-(x-1),4(x+y)-(2x+4)=8y

解9y=3x-(x-1)2x-9y+1=0①4(x+y)-(2x+4)=8y2x-4y-4=0②②-①得:5y-5=0∴y=1将y=1代入①得:2x-9+1=0∴x=4∴方程的解为:x=4,y=1

已知x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y

1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=

若|x+2y-1|+y²+4y+4=0,求(2x-y)²-2(2x-y)(x+2y)+(x+2y)&

∵|x+2y-1|+y²+4y+4=0∴|x+2y-1|+(y+2)²=0∴x=5,y=-2(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x

[(y-2x)(-2x-y)-4(x-2y)²]*2y,其中x=1,y=2

x)(-2x-y)-4(x-2y)²]*2y,其

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2