x²-4y² 35 xy • 28x² x² -4xy 4y2

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因式分解:x³-5x²y-24xy²

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

如果x^2+xy+y-14,y^@+xy+x=28,求x+y的值.

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

x²-6xy+9y²???因式分解

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

(xy-x^2)乘以(xy)/(x-y)

对.前提是x不等于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?

(-3x^y+2xy)-( )=4x^+xy

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

计算:4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2

4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2=(4xy-x²-2xy-y²)/[xy(x+y)(x-y)]÷[(x-y)(x+2y

计算:(xy-x²)×x-y/xy

这题少括号了吧.

已知xy≠0,且x²-3xy-4y²=0,则x比y的值为多少?

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

已知x²+9y²+4x-6y+5=0,求xy的值

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

x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2

x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2=16x^3y^3-49x^3y^3-45^3y^3=-78x^3y^3如果本题有什么不明白可以追问,

(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)

(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)=1.=3(a-b)/(10ab)*(25a^2b^3)/((a+b)(a-b))=15/2*(ab^2)/(a+

已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

已知xy大于0求证xy+1/xy+y/x+x/y大于等于4

xy+1/xy+y/x+x/y=[(xy)^2+1+x^2+y^2]/(xy)=[(xy)^2-2xy+1+x^2-2xy+y^2+4xy]/(xy)=[(xy-1)^2+(x-y)^2+4xy]/(

已知x+y=0,x+13y=1,求x²+12xy+13y²的值.

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+

已知x-y=4xy,则2x+3xy-2yx-2xy-y

∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.

在实数范围内因式分解:1 )3x²-8xy+3y² 2 )5x²y²+xy-73 )3x∧4+5x²-2

解题思路:先用十字相乘法、再运用平方差公式可解。解题过程:同学:另两道题目,我一时间没能解答出来,请你再检查一下原题。很抱歉,请你原谅!

x^2y+4xy+4y

x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²

{x-y+(4xy/x-y)}{x+y-(4xy/x+y)}怎么解?

原式={[(x-y)^2+4xy]/(x-y)}{[(x+y)^2-4xy]/x+y}={[x^2-2xy+y^2+4xy]/(x-y)}{[x^2+2xy+y^2-4xy]/x+y}={[x^2+2