m、n满足|m+2|+根号n-4=0,分解因式(x²+y²)-(mxy+n)

来源:学生作业帮助网 编辑:作业帮 时间:2024/06/09 22:13:16
m、n满足|m+2|+根号n-4=0,分解因式(x²+y²)-(mxy+n)
xՒOK0ƿJA6CAE C/+0 3!^A øyW}tfv$$yyV~̓) HN%[@RtJ͍=ѦnuS7UeR7X+]6aCxG=9^M35Ns07 x %NYD8.fd؝ + ~vd@@bP"S1ўgg梌a;}HN:P}t#-lee\X\d4U:Dk)DLAE#/vW+zJp:i0r&

m、n满足|m+2|+根号n-4=0,分解因式(x²+y²)-(mxy+n)
m、n满足|m+2|+根号n-4=0,分解因式(x²+y²)-(mxy+n)

m、n满足|m+2|+根号n-4=0,分解因式(x²+y²)-(mxy+n)
|m+2|+根号n-4=0
则必有:
m+2=0
n-4=0
解得:m=-2;n=4
(x²+y²)-(mxy+n)
=(x²+y²)-(-2xy+4)
=x²+2xy+y²-4
=(x+y)²-2²
=(x+y+2)(x+y-2)

因为:|m+2|+根号n-4=0,则:
m+2=0且n-4=0,得:m=-2、n=4
(x²+y²)-(mxy+n)
=x²+y²+2xy-4
=(x+y)²-4
=(x+y+2)(x+y-2)

|m+2|+根号n-4=0
由题意得
m+2=0
n-4=0
解得 m=-2 n=4
将m n值代入
(x²+y²)-(mxy+n)
=(x²+y²)-(-2xy+4)
=(x+y)²-4
=(x+y-2)(x+y+2)

|m+2|+根号n-4=0
则必有:
m+2=0
n-4=0
解得:m=-2;n=4
(x²+y²)-(mxy+n)
=(x²+y²)-(-2xy+4)
=x²+2xy+y²-4
=(x+y)²-2²
=(x+y+2)(x+y-2