"(1 3x y)(1 9x² y²)(1 3x-y)"计算-搜答案-搜搜考试网

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

解题思路：提取公因式进行分解解题过程：附件最终答案：略

x²/xy-x/y

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

第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0（a-4)(a+5)=0得

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

解题思路：本题主要利用完全平方公式进行因式分解即可求出结果解题过程：解:x²－6xy＋9y²=(x-3y)2

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-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?

=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²

这题少括号了吧.

解题思路：将原方程右边分解因式为（x-4y)(x+y)=0,可得x=4y或x=-y,进而可求出x与y的比值。解题过程：解：∵x2-3xy-4y2=0∴（x-4y)(x+y)=0∴x=4y或x=-y∵x

解题思路：根据题意，由完全平方公式的知识可求解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inc

【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

x^2-10x-11=(x-10)(x+1)x^2y^2-16xy+60=(xy-10)(xy-6)(x+y)^2-7(x+y)-18=(x+y-9)(x+y+2)x^2-13xy-30y^2=(x-

x^2+xy+2y^2=9+y2x^2+2xy+4y^2=18+2y2x^2+2xy+4y^2+x=1918+2y+x=19x+2y=1x=1-2y(1-2y)^2+(1-2y)y+2y^2-y=94

x²+y²=19(x+y)²=x²+y²+2xy=25xy=3(x-y)*(x-y)=(x-y)²=x²+y²-2xy=

原式=x²y²/(x-y)^4×[-y³(x-y)³]/x³=-y^5/[x(x-y)]=-y^5/(x²-xy)

解题思路：：∵x+y=0,x+13y=1，解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程：已知x+

x²+xy=12 xy+y²=13

两式相加得到x+y=5,相减得y-x=1/5,故x=12/5,y=13/5xy=156/25,因为要求的都是正数,而且xy同正负,所以只考虑x,y正数即可故x²+y²=（x+y）^