(2 3x-y)(2 3x y)
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(x+1)(y+1)
x/(xy-y²)=x/[y(x-y)]y/(x²-xy)=y/[x(x-y)]∴公分母为:xy(x-y)通分后得x/(xy-y²)=x²/[xy(x-y)]y
6x²-5xy-6y²+2x+23y-20=(3x+2y)(2x-3y)+2x+23y-20=(3x+2y-5)(2x-3y+4)备注:6x²-5xy-6y²这
设根号下x+y=a,则a平方+a=12可算出a1=-4,舍去a2=3所以x+y=9再和x+xy+y=23组成方程组,解得x1=7,y1=2x2=7y2=7
求数学达人来解答x²/xy-x/y=(1-x)/y3x/4x+y-x-2y/4x+y=(3x-x-2y)/(4x+y)(1-y/y+x)÷x/y²-x²=(1-y)*(y
设t=根号(x+y)t>=0所以t^2+t-12=0得出(t-3)(t+4)=0因为t>=0,所以t=3所以x+y=9带入二式可得9+xy=23,xy=14联立x+y=9,xy=14得出x(9-x)=
XY(XY-X^2)(X-Y)=X^2Y(X-Y)(X-Y)=(X-Y)(X^2Y-1)再问:答案是。-X2Y.我不知道过程是如何来的。能告诉我过程吗。谢谢。负X的平方Y
对.前提是x不等于y
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
xy-1+x-y=XY+X-Y-1(加法交换律)=(XY+X)-(Y+1)=X(Y+1)-(Y+1)(提取公因式X)=(Y+1)(X-1)(提取公因式Y+1)这样可以么?
2x²+6y²+23xy=(X²+7xy)乘以2+(2y²+3xy)乘以3+后面缺什么补什么(这个不用补罢了)这样的题都可以这么想是一种方法要会答案自己去做我告
2x²+6y²+23xy=2(x²+7xy)+3(2y²+3xy)=32*2+46*3=202
题目抄错了吧?是求2x²+6y²+23xy的值由已知可得:2x²+14xy=646y²+9xy=138上面两式相加得:2x²+6y²+23x
=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²
这题少括号了吧.
【x²+xy/(x-y)】/【xy/(x-y)】=【x²(x-y)/(x-y)+xy/(x-y)】/【xy/(x-y)】={【x²(x-y)+xy】/(x-y)}/【xy
若是209,则xy=8,x+y=15,算出x,y就不是整数了,与题意不符.若是34,x,y为3,5,符合题意.
原式=x²y²/(x-y)^4×[-y³(x-y)³]/x³=-y^5/[x(x-y)]=-y^5/(x²-xy)
{23x+17y=63……①17x+23y=57……②①+②得40x+40y=120即:x+y=3……③①-②得6x-6y=6即:x-y=1……④③+④得,2x=4,所以x=2把x=2代入③,得2+y
xy+x+y=23,x²y+xy²=120,xy(x+y)=120把xy,x+y看成是z²-23z+120=0的两根解得z1=15,z2=8又把x,y看成是m²