解方程组6(x-y)-7(x y)=21

第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0（a-4)(a+5)=0得

将y=x-3代入x^2-xy-6=0得x^2-x(x-3)-6=3x-6=0从而得x=2和y=-1

xy/(2x+y)-xy/(2x-y)=3xy/(2x-y)+xy/(2x+y)=4；一、两公式对加得2xy/(2x+y)=7xy=(2x+y)*7/2二、把xy代入原公式求解就行了：(2x+y)*7

XY=15X+Y=8Y=8-XX(8-X)=15-X^2+8X-15=0X^2-8X+15=0(X-3)(X-5)=0所以X=3或X=5得Y=5或Y=3

x+y=7-xyx^2+y^2=10即(x+y)^2=(7-xy)^2=x^2+y^2+2xy=10+2xy即(7-xy)^2=10+2xy即x^2y^2+39-16xy=0解得xy=3代入x+xy+

消掉y得2x(x-1)=2即x(x-1)=1,所以,x=(1+根号5)/2,对应y=（根号5-1）/2或x=(1-根号5)/2,对应y=（-根号5-1）/2这里用了求根公式!得解!

xy\X+Y=12\71/y+1/x=7/12(1)YZ\Y+Z=6\51/z+1/y=5/6(2)XZ\X+Z=4\31/z+1/x=3/4(3)由(1)-(2)得1/x-1/z=-1/4(4)由(

(x²+y²)(x²-y²)=15xy(x²-y²()-=6相除2x²+2y²=5xy(x-2y)(2x-y)=0x=2

(1)×7+(2)×328y+27y=55xy所以y=0或x=1分别代入所以x=0.y=0x=1,y=1/2

xy/3x+2y=1/8,xy/2x+3y=1/73x+2y/xy=8,2x+3y/xy=73/y+2/x=8,2/y+3/x=73/y+2/x=8,2/y+3/x=71/x=1,1/y=2,x=1,

x+y=7x=7-y(7-y)y=127y-y^2=12y^2-7y+12=0(y-3)(y-4)=0y1=3y2=4x1=4x2=3

两式相加得x=4+xy即x=4/(1-y)①两式相减得y=-1-4xy②①代入②y=-1-4y*4/(1-y)整理得y²-16y-1=0解得y=8±√65代入①,即可算出x了如有不明白,可以

x=2,y=0.5

啊,由于我只是初二,能力有限,所以我只知道猜的这种方法由x+y=10得y=10-x把y=10-x代入XY=24,得(10-x)x=24由(10-x)x=24得x²-10x+24=0用十字相乘

第一个,由①x+3y=7得,(x+3y)(x-3y)=x²-9y²=7(x-3y)而由2知x²-9y²=-35所以(x-3y)=-5联立①x+3y=7得x=1,

因为：xy=(x-10)(y+1)→xy=xy+x-10y-10①xy=(x-15)(y+3)→xy=+3x-15y-45②由①、②得：x=10y+10③x=5y+15④将③‐­④得：10y+

∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy

由(x-2)(y+3)=xy(x-5)(y+2)=xy得3x-2y-6=02x-5y-10=0上面方程化为6x-4y-12=06x-15y-30=0两式相减得11y=-18,y=-18/11,代入任意

x=oy=o再问：错了再答：哪里错？再答：验证符合哎再问：x+y=xz,x+2y=xz²再问：x,y≠0再答：你都给的错题再问：呵呵再答：算我对啦😜再问：额

x1=-1/12*(1261+234*29^(1/2))^(1/3)-13/12/(1261+234*29^(1/2))^(1/3)+5/12x2=1/24*(1261+234*29^(1/2))^(