圆x²+y²-2x-8y+13=0的距离为一,则a=
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(x+2y)(x-2y)-(2y-x)(-2x-y)=(x+2y)(x-2y)+(x-2y)(-2x-y)=(x-2y)[(x+2y)+(-2x-y)]=(x-2y)(-x+y)代入x=8y=-8=-
[(x+2y)(x-2y)-(x-2y)^2+8y(x+y)]/4x=[x^2-4y^2-x^2-4y^2+4xy+8xy+8y^2]/4x=12xy/4x=3y
8(x平方-2y平方)-x(7x+y)+xy=8x平方-16y平方-7x平方-xy+xy=x平方-4y平方=(x-2y)(x+2y)手机提问的朋友在客户端右上角评价点【满意】即可.再问:太棒了!连我这
-x^y^7+y^8)(这里有问题,x的几次方?再问:就是一次方再答:(x^8-x^7y+x^5y^3-x^4y^4+x^3y^5-xy^7+y^8)(x^2+xy+y^2)=(x^8-x^7y+x^
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
【(x-y)^2+(x+y)(x-y)】除以2x=(x-y)*(x-y+x+y)/2x=(x-y)*2x/2x=x-y
2x+y+z=10(1)x+2y+z=-6(2)x+y+z=8(3)(2)-(3):y=-14(4)(1)-(2):x-y=16(5)把(4)代入(5):x+14=16x=2(6)把(4)和(6)代入
=(x+y)(6-16)+(x-y)(2-5-1)=-10(x+y)-4(x-y)再问:能讲过程吗?谢谢!再答:先把(x+y)提出来,在把(x-y)提出来
(1){x=y-3{x+y=11y-3+y=112y=14y=7x=y-3=7-3=4所以原方程组的解为x=4,y=7(2){2x+y=13{3x-2y=8y=13-2x3x-2(13-2x)=83x
(x-y)^2+7(x+y)-11(x-y)^2-8(x+y),=-10(x-y)²-(x+y)当x+y=-3,x-y=-1时=-10*(-1)²-(-3)=-10+3=-7
(1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/
区域是一个正方形
1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8y^7/(x^8+y^8)=【(x+y)-(x-y)】/(x-y)(x+y)-2y/(x^2+y^2)-4y^
解题思路:由题意x²+y²-4x-2y+8=(x-2)²+(y-1)²+3,然后根据完全平方式的性质进行求解.解题过程:解:∵x2+y2-4x-2y+8=(x2-4x+4)+y2-2y+1+3=(x-
原式=(5-8)(x-y)^2-(1+3)(x-y)=-3(x-y)^2-4(x-y)
1.((y+2x)(-2x+y)+4(x+2y)^2)/3y=(y^2-4x^2+4x^2+16xy+16y^2)/3y=(17y^2+16xy)/3y=17y/3+16x/32.(2(x-y)^3-
x+y/y=11/8说明11y=8(x+y)即8x=3y所以x-y/x+2y=3(x-y)/3(x+2y)=(3x-3y)/(3x+6y)=(3x-8x)/(3x+16x)=-5/19
等于.-3y-10x