初中二年级下数学题
{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分之x²+2xy+y²}×{x&...
{x²-(y-z)²分之x²-y²}÷{(x-y)²-z²分之x²+2xy+y²}×{x²-xy分之x²+xy-xz}
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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
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
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解 (x+y)(x-y) ( x-y+z)(x-y-z) x(x+y-z)
= ──────────×─────────× ───────
(x+y-z)(x+z-y) ( x+y)² x(x-y)
x-y-z
=───────
x+y
= ──────────×─────────× ───────
(x+y-z)(x+z-y) ( x+y)² x(x-y)
x-y-z
=───────
x+y
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解 (x+y)(x-y) ( x-y+z)(x-y-z) x(x+y-z)
= ──────────×─────────× ───────
(x+y-z)(x+z-y) ( x+y)² x(x-y)
x-y-z
=───────
x+y
= ──────────×─────────× ───────
(x+y-z)(x+z-y) ( x+y)² x(x-y)
x-y-z
=───────
x+y
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