3x 2y=m x 4y=4 的解为非负数 求m取值范围
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(x+y)(xy)=x^2y+xy^2=-8原式=-7
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.
∵x+y=6,xy=4,∴x2y+xy2=xy(x+y)=4×6=24.故答案为:24.
原式=4x2y-6xy+3(4xy-2)+x2y+1=5x2y+6xy-5当x=2,y=-12时,原式=5×4×(-12)+6×2×(-12)-5=-21.
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
∵2x+y=4,xy=3,∴2x2y+xy2=xy(2x+y)=3×4=12.故答案为:12
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
题目1看不明白解题目2x+y=4,(x+y)^2=4^2=16,同样x-y=10,(x-y)^2=10^2=100,(x+y)^2=x^2+2xy+y^2,(x-y)^2=x^2-2xy+y^2,(x
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+y=5,xy=6,∴x2y+xy2=xy(x+y)=5×6=30.故答案为:30.
根据题意得,算式为:-3x2-[-4x2y+(-5x2)+2x2y],-3x2-[-4x2y+(-5x2)+2x2y]=-3x2+4x2y+5x2-2x2y=2x2+2x2y=2x2(1+y).
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
x=±1,y=±3,z=±2xyzz>y则0>x>z>yx=-1,y=-3,z=-2,x2y-[4x2y-(xyz-x2z)-3x2z]-2xyx=x2y-4x2y+xyz-x2z+3x2z-2xyx
代入x=-1,y=1,2x^y-(5xy^-3x^y)-x^=2*(-1)^*1-{5*(-1)*1^-3*(-1)^*1}-(-1)^=2-(-5-3)-1=9备注:2^表示2的平方
5x2y+3x2y+(-4x2y)=(5+3-4)x2y=4x2y,故答案为:4x2y.
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.
如果x,y符号相反,绝对值相等,即y=-x,代入原方程组,得3x-2x=m+1,4x-2x=m-1,即x=m+1,2x=m-1解之,2(m+1)=m-1,得m=-3如果x比y大1,即x=y+1,代入原
由柯西-黎曼条件v'(x)=-u'(y),v'(y)=u'(x)得u'(y)=-6xy,u'(x)=3y²-3x²因而选择B
5x2y+3x2y2+(-4xy2)=5x2y+3x2y2-4xy2.