已知x y 2的绝对值 (2x-3y 1)²=0求xy的值
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x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
∵mx3+3nxy2-2x3-xy2+y=(m-2)x3+(3n-1)xy2+y,多项式中不含三次项,∴m-2=0,且3n-1=0,解得:m=2,n=13,则2m+3n=4+1=5.故选D
①x2y+xy2=xy(x+y)=1×3=3;②x2+y2=(x+y)2-2xy=32-2×1=7.
绝对值x-2+绝对值4-y等于0则x-2=04-y=0则x=2y=43x+2y=3×2+2×4=14再问:不是等于3吗再答:怎么等于3??按你的条件就能求出这个结果的
由题意得,x-1=0,y+3=0,解得x=1,y=-3,所以,1-xy-xy2=1-1×(-3)-1×(-3)2,=1+3-9,=4-9,=-5.
原式=(4x2y+5xy2+3x-2y+5)-2(2x2y-3xy2-2x+1)=4x2y+5xy2+3x-2y+5-4x2y+6xy2+4x-2=11xy2+7x-2y+3.
2(xy-5xy2)-(3xy2-xy)=(2xy-10xy2)-(3xy2-xy)=2xy-10xy2-3xy2+xy=(2xy+xy)+(-3xy2-10xy2)=3xy-13xy2,∵(x+1)
已知:关于x,y的多项式2mx3+3nxy2-2x3-xy2-2x+y中不含三次项,所以2m-2=0;3n-1=0;所以m=1;n=1/3;所以2m+3n的值=2×1+3×1/3=2+1=3;很高兴为
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
3x-4y+2的绝对值+x-2y的绝对值=0则:3x-4y+2=0x-2y=0解得:x=-2;y=-1
xy>0则x和y同号所以x=2,y=3或x=-2,y=-3所以x+y=5或-5|x+y|=5x-y=-1或1|x-y|=1所以原式=5-1=4再问:你确定再答:嗯
1)-x^2y-xy^2=-xy(x+y)=-2×3=-62)x²+5xy+30y=x(x+5y)+30y=6x+30y=6(x+5y)=6×6=36
1=x+2y=x+y+y≥3(xy²)^(1/3).===>xy²≤1/27.等号仅当x=y=1/3时取得,∴(xy²)max=1/27
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 /-2/=2 y不用变 Z=0解原式是x2-[4xy2-(xyz-x2z0-3x2z]-2xyz=4-
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²
mx3+3nxy2+2x2-xy2+y=mx3+(3n-1)xy2+2x2+y不含三次项式所以m=03n-1=0n=1/3因而2m+3n=1不懂再问哦
(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
原式=-xy(x-y),当x-y=3,xy=-2时,则原式=-3×(-2)=6.故答案为:6.
解3xy²-[2xy²-2(xy-1.5x²y)]+xy-3x²y=3xy²-(2xy²-2xy+3x²y)+xy-3x²