{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/05 10:32:18
{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分之x²+2xy+y²}×{x²-xy分之x²+xy-xz}
楼主题弄错了吧?以后打分数就打 / 大多数人都能看懂,还有的重视运算的法则.若是觉得打起来容易让人难理解,就照下来再传上来.只要像素还可以其实200万都够了.
若是{(x²-y²)]/[x²-(y-z)²]}÷{(x²+2xy+y²)/[(x-y)²-z²]}×[(x²+xy-xz)/(x²-xy)]
={(x²-y²)]/[x²-(y-z)²]}•{[(x-y)²-z²]/(x²+2xy+y²)}•[(x²+xy-xz)/(x²-xy)]
={(x²-y²)][(x-y)²-z²](x²+xy-xz)}÷{[x²-(y-z)²](x²+2xy+y²)(x²-xy)]}
=[(x-y)(x+y)(x-y-z)(x-y+z)(x+y-z)x]÷[(x-y+z)(x+y-z)(x+y)²(x-y)x]
=(x-y-z)/(x+y)
若是[x²-(x²-y²)/(y-z)²]÷[(x-y)²-(x²+2xy+y²)/z²]×[x²-(x²+xy-xz)/(xy)]只能消掉其中的一部分,没多大意义
若是[x²-x²/(y-z)²-y²]÷[(x-y)²-x²/z²+2xy+y²]×[x²-x²/(xy)+xy-xZ]还是只能消掉其中的一部分,没多大意义
再问: 确定?
再答: 确定啊,你传张照片上来吧,那样不会把题意弄错
若是{(x²-y²)]/[x²-(y-z)²]}÷{(x²+2xy+y²)/[(x-y)²-z²]}×[(x²+xy-xz)/(x²-xy)]
={(x²-y²)]/[x²-(y-z)²]}•{[(x-y)²-z²]/(x²+2xy+y²)}•[(x²+xy-xz)/(x²-xy)]
={(x²-y²)][(x-y)²-z²](x²+xy-xz)}÷{[x²-(y-z)²](x²+2xy+y²)(x²-xy)]}
=[(x-y)(x+y)(x-y-z)(x-y+z)(x+y-z)x]÷[(x-y+z)(x+y-z)(x+y)²(x-y)x]
=(x-y-z)/(x+y)
若是[x²-(x²-y²)/(y-z)²]÷[(x-y)²-(x²+2xy+y²)/z²]×[x²-(x²+xy-xz)/(xy)]只能消掉其中的一部分,没多大意义
若是[x²-x²/(y-z)²-y²]÷[(x-y)²-x²/z²+2xy+y²]×[x²-x²/(xy)+xy-xZ]还是只能消掉其中的一部分,没多大意义
再问: 确定?
再答: 确定啊,你传张照片上来吧,那样不会把题意弄错
{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分
已知x+y+z=0,求(y²+z²-x²)分之1+(z²+x²-y
若x-y=6,xy=-8,求代数式(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)的值
25(x-y)²-10(y-z)+1因式分解
x³+y³+z³-3xyz为什么等于(x+y+z)(x²+y²+z
(y³z分之-3x)²
(x-3y+2z)²=
x≠y≠z,x+y分之一=y+z分之一=z+x分之一,求x²y²z²
已知x²+y²+z²-2x+4y-6z+14=0,求x+y-z/x+y+z的值
已知x,y,z满足x+y+2z=1,x²+y²+6z+1.5=0,求x,y,z的值
已知丨X+2丨+(Y-3)² =0 且(5分之X-2Y+Z)+5=½Y+X+Z 求Z
1、x³+x²y-x²z-xyz