已知|x y-4| x-y 10的算术平方根
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原式=3xy+6y-xy+6x=2xy+6(x+y)=2*(-2)+6*4=24-4=20
因为x-y=4xy所以x-2xy-y=2xy2x+3xy-2y=11xyx-2xy-y分之2x+3xy-2y=5.5
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15
原式=6x+4xy+2xy+3x+2y-4xy+7y=9x+9y+2xy=9(x+y)+2xy=9×4+2×2=36+4=40
5xy+4x+7y+6x-3xy-4xy+3y=5xy-3xy-4xy+4x+6x+7y+3y=(5-3-4)xy+(4+6)x+(7+3)y=-2xy+10x+10y=-2xy+10(x+y)因为x
原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1
题目打错了"则f(x,y)-f(X1,Y10=O"
∑xi=20,x'=20/5=4∑yi=66,y'=66/5=13.2∑xi^2=90∑xiyi=282b=(n∑xiyi-∑xi∑yi)/[n∑xi^2-(∑xi)^2]=(5*282-20*66)
先把代数式(5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)化简(5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)=5XY+4X+7Y+6X-3XY-4XY+3Y=-2XY+10X+1
a-2b=(5x²y-3xy²+4xy)-2(7xy²-2xy+x²y)=5x²y-3xy²+4xy-14xy²+4xy-2x
解题思路:先合并同类项,再结合已知条件变形,代入数值进行计算解题过程:求代数式(2x+3y-2xy)-(x+4y+xy)-(3xy+2y-2x)的值解:(2x+3y-2xy)-(x+4+xy)-(3x
解x²+2y²-2xy+4y+4=0吧(x²-2xy+y²)+(y²+4y+4)=0(x-y)²+(y+2)²=0∴x-y=0,y
|x-y|=4,xy=-3(x+y)^2=(x-y)^2+4xy=14-12=2x+y=根2,-根2xy^2+x^2y=xy(x+y)=-3根2,3根2再问:(x+y)^2=(x-y)^2+4xy=1
因为x、y都是正数,则:x+4y≥4√(xy)设:√(xy)=t,则:xy=4y+x+5≥4√(xy)+5即:t²≥4t+5t²-4t-5≥0t≤-1或t≥5因为:t=√(xy)≥
∵x-y=4xy∴原式=[2(x-y)+3xy]/[-(x-y)-2xy]=(8xy+3xy)/(-4xy-2xy)=11xy/(-6xy)=-11/6
密度按7.8克/立方厘米计算.
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采