满足方程3x 2y=k 1,2x 3y=5
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![满足方程3x 2y=k 1,2x 3y=5](/uploads/image/f/5969567-47-7.jpg?t=%E6%BB%A1%E8%B6%B3%E6%96%B9%E7%A8%8B3x+2y%3Dk+1%2C2x+3y%3D5)
原式=x3-2y3-3x2y-3x3+3y3+7x2y=-2x3+y3+4x2y
y1=k1/3,y2=k2/0.5,y1=-y2,k1/3=-k2/0.5,k1/k2=-3/0.5k1/k2=-6
因为A+B+C=x3-2y3+3x2y+xy2-3xy+4+y3-x3-4x2y-3xy-3xy2+3+y3+x2y+2xy2+6xy-6=1,所以,对于x、y、z的任何值A+B+C是常数.
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
∵代数式x3+y3+3x2y+axy2含有因式x-y,∴当x=y时,x3+y3+3x2y+axy2=0,∴令x=y,即x3+x3+3x3+ax3=0,则有5+a=0,解得a=-5.将a=-5代入x3+
(1)原式=-(a-1)2;(2)原式=-x(x-y)2;(3)原式=(x+1)2(x-1)2;(4)原式=(x+y-x+y)2=4y2.
化简得:9-12Y^2+6Y+4+12Y^2+4Y-10-10Y+X-Y+1=X-Y+4带入X、Y值得:=3
(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=-2y3=-2×(-1)3=2.因为化简的
A+B+C=(x3+3x2y-5xy2+6y3-1)+(y3+2xy2+x2y-2x3+2)+(x3-4x2y+3xy2-7y3+1)=(1+1-2)x3+(3+1-4)x2y+(-5+2+3)xy2
|x-2|+(y+3)²=0都是非负式所以分别都=0所以x-2=0y+3=0所以x=2y=-3又因为z是最大的负整数所以z=-1原式=2(x²y+xyz)-3(x²y-x
根据题意得:(x3-3x2y)-(3x2y-3xy2)=x3-3x2y-3x2y+3xy2=x3-6x2y+3xy2,故选C.
原式=x3+3x2y-5xy2+6x3+1-2x3+y3+2xy2+x2y+2-4x2y-7x3-y3+4xy2+1=-2x3+xy2+4,由于y为偶次幂,故误把“x=3,y=-1”写成“x=3,y=
(1)(x3-2x2y+3y2)-(-2x3-3x2y+5y2)=x3-2x2y+3y2+2x3+3x2y-5y2=3x3+x2y-2y2,答:这个多项式为3x3+x2y-2y2.(2)当x=-12,
答案:2x^2y+2xy^2原式=4x2y-{x2y-[3xy2-2x2y+4xy2+x2y]}-5xy2=4x2y-{x2y-[7xy2-x2y]}-5xy2=4x2y-{x2y-7xy+x2y]}
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
多项式3x2y-5xy3+y2-2x3的各项为3x2y,-5xy3,y2,-2x3,按x的降幂排列为-2x3+3x2y-5xy3+y2.故答案为:-2x3+3x2y-5xy3+y2.再问:为什么是这样
原式=2x2y+2xy-3x2y+3xy-4x2y=-5x2y+5xy,当x=-1,y=1时,原式=-5×(-1)2×1+5×(-1)×1=-5-5=-10.
2k1+k2/2=3①-k1-k2=-3②①×2+②得到3k1=3解得k1=1代入②解得k2=2
x3-y3-x2y+xy2=(x-y)(x2+xy+y2)-xy(x-y)=(x-y)(x2+xy+y2-xy)=(x-y)(x2+y2)
楼上的想法比较正确,但是有错误,利用隔板法在12个空隙中插3个板,运用C(12,3)这样做忽略了两个板插在一个空隙里的情况.比如(0,1,2,3)这组解,利用这种算法就是求不出的.就是说,如果用组合算