-24x²y-12xy² 28y³

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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

3x 的2y次方+28y的平方+7xy的平方+12xy因式分解

原式=(3x²y+12xy)+(7xy²+28y²)=3xy(x+4)+7y²(x+4)=y(x+4)(3x+7y)

若x²+xy+y=14,y²+xy+y=28,求x+y的值

x^2+xy+y=14y^2+xy+x=28+x^2+2xy+y^2+x+y=42(x+y)^2+(x+y)=42(x+y)(x+y+1)=0则x+y=0或x+y=-1

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?

已知X+y=10,XY=24,求(X-Y)(X-Y)的值

X+y=10,XY=24,则X=4,Y=6或X=6,Y=4,所以(X-Y)(X-Y)=2*2=4或(-2)*(-2)=4,都为4

计算:(xy-x²)×x-y/xy

这题少括号了吧.

x,y都是自然数,且x(x-y)-y(y-x)=12,求x+y-xy的值

x(x-y)-y(y-x)=12那么:化简得:(x+y)*(x—y)=12x、y是自然数,所以x1=2,y1=4x2=4,y2=2所以x+y-xy=-2

(x^2+xy/x-y)/(xy/x-y)计算

【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

16x³-24x²y²+12xy^4-2y^6

16x³-24x²y²+12xy^4-2y^6=(16x^3-16x^2y^2+4xy^4)-8x^2y^2+8xy^4-2y^6=4x(4x^2-4xy^2+y^4)-

xy=-2,x+y=3,求代数式(4xy+12y)+[7x-(3xy+4y-x)]

(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

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+

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)^

4x^2y+12xy+9y

4x^2y+12xy+9y=y(4x^2+12x+9)=y(2x+3)^2

-24x的2次方y-12xy的2次方+28y的3次方 是分解因式 (x-y)²-(y-x) 也是因式分解 ,

-24x^2y-12xy^2+28y^3解,得:==-4y(6x^2+3xy-7y^2)分解因式(x-y)²-(y-x)也是因式分解,解,得:==(x-y)^2+(x-y)==(x-y)[(

若12x*x+12y*y=25xy,试求x/y的值

2边同时除以y*y,得到12(x/y)^2+12=25(x/y).然后算出就可以了.