2x+2y+2c=10 2x+2y+3c=7 x+3y+2c=6 x=?y=?c=?

a(x-y)(x+y)-b(y-x)(x+y)-c(x-y)^2(y+x)提取公因式 如果A={（x,y）|x,y+2>0},B={（x,y）|2x+3y-6>0},C={（x,y）|x-4 曲线C：2x^+y^-x+2y-3=0关于y轴对称 a/x-y,b/3y-3x,c/x^2-2xy+y^2通分 （x+y）^2-z^=a^2-(b-c)^2=(x-y)^3-x+y=16(x+y)^2-25(x-y)^2= 设集合A={x|x=y^2-2y-8},B={x|y=-x^2+2x+3},C={y|y+1 解微分方程y`=1/(x-y)+1答案(x-y)^2+2x=c 已知x/y+z=y/x+z=z/x+y,分式(x+2y+3z)/(2x+5y-2c)有意义时的值 C语言中printf((%d%d,x,y,(x,y))x=1,y=2 已知A=3x^2-12,B=5x^2*y^3+10x*y^3,C=(x+1)(x+3)+1求其公因式已知A=3x*x-12,B=5x*x*y*y*y+10x*y*y*y,C=(x+1)(x+3)+1求其公因式,若无,说出理由 2x+2y+2c=10 2x+2y+3c=7 x+3y+2c=6 x=?y=?c=? x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy 是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy 设A={(x,y)}|3x+2y|.设A={(x,y)}|3x+2y|.B={(x,y)|x-y=2|},C={(x,y)|2x-2y|=3},D={(x,y)|6x+4y=2|},求A∩B,B∩C,A∩D设A={(x,y)|3x+2y=1}.B={(x,y)|x-y=2},C={(x,y)|2x-2y=3},D={(x,y)|6x+4y=2},求A∩B，B∩C，A∩D 已知2y-x=5,则5(x-2y)方-3(x-2y)-60的值是?A.80 B.10 C.210 C.40 几道 数学题急啊 对了再加20分(x+y)^2-(x+y)^3(x-y)^2-(x-y)^3 (x-y)^2-(y-x)^32(x-y)-3(y-x)^24ab(a+b)^2-6a^2b(a+b)(x+y)(x-y)-(x+y)^28a(x-y)^2-4b(y-x)a=-5 a+b+c=-5.2求代数式a^2(-b-c)-3.2a(c+b)的值 分解因式1.x^2+y^2+2x+6y+10 2.(1-y)^3+4(y-x) 3.(a-c)^2-4(a-b)(b-c)