求方程x y=x2-xy y2整数解
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XY+X+Y-7=0xy+x+y+1=8(x+1)(y+1)=8=1*8=2*4=-1*(-8)=-2*(-4)x+1=1,y+1=8,得:x=0,y=7x+1=8,y+1=8,得:x=7,y=0x+
xy+y+2x+2-2=0y(x+1)+2(x+1)=2(y+2)(x+1)=2iy+2=1,x+1=2得出y=-1,x=1iiy+2=2,x+1=1得出y=0,x=0iiiy+2=-1,x+1=-2
|xy|-|2x|+|y|=4==>|x||y|-2|x|+|y|=4==>|x||y|+|y|-2|x|-2=2==>(|y|-2)(|x|+1)=2|x|+1为正整数|y|-2也为正整数2分解质因
xy-y=xy=x/(x-1)是整数因为相邻的两个整数互质互质,又是倍数关系所以只有0÷(-1)和2÷1所以x-1=-1,x-1=1x=0,x=2所以x=0,y=0x=2,y=2
x=1x=-1y=3y=-34种组合
2x^2+3xy-2y^2-3=(2x-y)(x+2y)-3=0(1)2x-y=1,x+2y=3x=1,y=1(2)2x-y=3,x+2y=1非整数解,舍去(3)2x-y=-1,x+2y=-3x=-1
(21-14√3)/9=-0.36(21+14√3)/9=5.03∴y的取值为012345再问:怎样解27y^2-126y-49≤0这个不等式?谢谢。再答:好像不能因式分解只有用(-b-√b^2-4a
(x-2)(y-2)=7x,y是整数x-2,y-2是整数x-2=1,y-2=7x-2=7,y-2=1x-2=-1,y-2=-7x-2=-7,y-2=-1x=3,y=9x=9,y=3x=1,y=-5x=
正确的解法如下:将已知变形为:xy+3x-y=20x(y+3)-(y+3)=17(x-1)(y+3)=17=1×17=-1×(-17)所以有四种情形:①(x-1)=1(y+3)=17得:x=2,y=1
1.将式子化成y=(4x)/(3x-1)的形式,从0套起,得出所有整数解是:x=0,y=0x=1,y=2x=-1,y=12.首先,0是自然数,若n不等于0,则有:假设n2+n+1是某个整数的平方,有n
xy-x+y=0xy-x+y-1=-1x(y-1)+(y-1)=-1(x+1)(y-1)=-1-1=1*(-1)所以x+1=1,y-1=-1或x+1=-1,y-1=1所以x=0,y=0x=-2,y=2
xy-2x-2y+7=0(x-2)(y-2)=-3x-2与y-2异号且x,y均为整数,-3的公因式组为-1×3,1×(-3)因为x≤y当x-2=1时,y-2=-3即x=3,y=-1(舍去)当x-2=-
∵x2-xy-5x+5y-1=0∴x(x-y)-5(x-y)=1∴(x-5)(x-y)=1,∵x,y为整数,则x-5为整数,x-y为整数,整数的乘积=1,则x-5=1,x-y=1或x-5=-1,x-y
x+y只能为整数.(1)若x+y≥4,则2xy=(x2+y2)(x+y-3)>0,只能x>0,y>0.此时,若x+y>4,则(x2+y2)(x+y-3)>x2+y2≥2xy,原方程无整数解.只能x+y
原式可变为x(1+y)=30-y,显然y不等于-1,所以x=(30-y)/(y+1)=31/(y+1)-1,所以y+1能整除31,因此y+1可取1,31,-1,-31,代入可得x分别等于30,0,-3
4种x=-3,y=4x=-1,y=4x=1,y=-4x=3,y=-4
设整数(x,y)满足方程x2-2xy+126y2-2009视为x的方程,判别式=4y2-4*(12y2-2009)=500(16-y2)+36应该是完全平方数且判别式>0所以y2再问:a2=ab+b2
由题意得XY-X-Y-4=0即(X-1)(Y-1)=5所以有X-1=5Y-1=1或X-1=1Y-1=5或X-1=-5Y-1=-1或X-1=-1Y-1=-5解得:X=6`````2`````-4````
(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2
7(x+y)=3(x^2-xy+y^2)=3[(x+y)^2-3xy]必有x+y=3k,k为整数因此7k=9k^2-3xy故也有k=3n-->x+y=9n,n为整数7n=27n^2-xy-->xy=2