(x²+y²-z²+2xy)/(x²-y²+z²-2xz)计算.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 18:35:48
xN@_{-q.Њ
6T}nU*"ġ*nxqulh?M3i|`M9gwe^mur~2>4x.)kjr}|3}19_
JZ0KxEUϬYż[s j̚A@Ui6[幚EϸAXmlociH6cJu*(jD+ώ~=}"iX?-֢^/y
p8npώn<%/b-p iB$ L!W)$BR2cQT
BAh+I^\x!Ņ eCBq*`{Ow}<=5U
(x²+y²-z²+2xy)/(x²-y²+z²-2xz)计算.
(x²+y²-z²+2xy)/(x²-y²+z²-2xz)计算.
(x²+y²-z²+2xy)/(x²-y²+z²-2xz)计算.
具体过程是这样的:
原式=[(x²+2xy+y²)-z²]/[(x²-2xz+z²)-y²)]
=[(x+y)²-z²]/[(x-z)²-y²]
=(x+y+z)(x+y-z)/(x-z+y)(x-z-y)
=(x+y+z)/(x-z-y)
(x+y-z)/(x-y+z)
(x+y)^2-Z^2/(x+z)^2-y^2=(x+y+z)(x+y-z)/(x-y+z)(x+y+z)=x+y-z/x-y+z
H
看看我的截图