6x^2-xy-2y^2-4x+5y-2因试分解

来源：学生作业帮助网 编辑：作业帮 时间：2024/09/30 09:14:17
6x^2-xy-2y^2-4x+5y-2因试分解
xTNA~bbA_ai}Tl HgA`R1UА wr+tv֫^tݝ3|3sqSYЋF64=QwG4O"BοwA/+\L!{I#1/F8.*v|ˤ FnL:BiU-PP88)/'mĺ J;,il J9Dtj*îvv?,]a&9ף&Ӱ܁ܠXGd4(@}&Zʗ"zGbGԘq&b\Ha&$o%MdR(:c8EifJ"> ѹvD8$s /YdO0]X="`*>+o>$^{''INҏ$I}sv)us1f)t0ivV$!9HyÑq S%Q$6S[E~֦iFZ\%D{4ݬrӚy\mم,( KDwPN% %|`#Oq4^p=,A%z{l~QC`*enjS7+h22~30Eƀ<a(GV74`CH:n`6x[ЋG30M[4Ǯ{ɮb纷VHh֮;^âoh2j\U`\3k˶V3

6x^2-xy-2y^2-4x+5y-2因试分解
6x^2-xy-2y^2-4x+5y-2因试分解

6x^2-xy-2y^2-4x+5y-2因试分解
原式=（3x-2y）(2x+y)-4x+5y-2 (十字相乘)
=(3x-2y+1)(2x+y-2)

请好评
如果你认可我的回答,敬请及时采纳,
~如果你认可我的回答,请及时点击【采纳为满意回答】按钮~
~手机提问的朋友在客户端右上角评价点【满意】即可.
~你的采纳是我前进的动力~~
O(∩_∩)O,记得好评和采纳,互相帮助

待定系数法结合十字交叉法

望采纳哈

6x^2-xy-2y^2-4x+5y-2
=(2x+y)(3x-2y)-4x+5y-2
下面我们用待定系数法
来解答
(2x+y)(3x-2y)-4x+5y-2=（2x+y+m)(3x-2y+n)
=6x^2-xy-2y^2-(3m+2n)x+(n-2m)y+mn
mn=-2
3m+2n=-4
n-2m=5
n=1
m...

全部展开

6x^2-xy-2y^2-4x+5y-2
=(2x+y)(3x-2y)-4x+5y-2
下面我们用待定系数法
来解答
(2x+y)(3x-2y)-4x+5y-2=（2x+y+m)(3x-2y+n)
=6x^2-xy-2y^2-(3m+2n)x+(n-2m)y+mn
mn=-2
3m+2n=-4
n-2m=5
n=1
m=-2
(2x+y)(3x+2y)-4x+5y-2=（2x+y-2)(3x-2y+1)
~如果你认可我的回答，请及时点击【采纳为满意回答】按钮~
~手机提问的朋友在客户端右上角评价点【满意】即可。
~你的采纳是我前进的动力~~
~~~~~~~~~~~！【来问吧】团队为您回答~~~~~~~~~~~~
O(∩_∩)O，互相帮助，祝共同进步！

收起

因式分解 x-xy+3y-3x 2x+xy-y-4x+5y-6 2X^+XY-Y^-4X+5Y-6 2x^+xy-y^-4x+5y-6=? 5x^3y*(-3y)^2+(-6xy)^2*(-xy)+xy^3*(-4x^2) 诺x+y=6,xy=4 求(5x-4y-3xy)-(8x-y+2xy) 化简：3x^2-[5xy-(x^y-3xy)+4x^2y-8xy] 因式分解 （3x+2y）-（x-y）² x²-xy+3y-3x 2x²+xy-y²-4x+5y-6 因式分解:12xy^2(y-x)+xy^3(x-y)+4x^2y(x-y) 因式分解 求过程 x-xy+3y-3x 2x+xy-y-4x+5y-6 3x平方-5x平方y-{2x平方y+6xy平方-[2xy平方-3x平方y+(3x平方y-4xy平方】} 4xy-[(x²+5xy-y²)-(3xy-1/2y)]x=3,y=-6 (x+y)(x+y+2xy)+(xy+1)(xy-1)(x+y)(x-y)+4(y-1) 计算:2xy/(x+y)².(x+y)/4y 2x^2-3xy+6y^2-x^2+4xy-5y^2 若(x*x+y*y)(x*x+y*y)-4x*x*y*y=0,求代数式(x*x+5xy+y*y)/(x*x+2xy+y*y)的值 3(x-y)²-7(x-y)+8(x-y)²+6(x-y)求详细过程（2） 5xy-4x²y-8xy²+3xy-xy²+4x²y x²-y²+5x+3y+4（x+y-2xy)(x+y-z)+(xy-1）² x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2