ax-bx+ay-by=··········化简
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ax-bx+ay-by=··········化简ax-bx+ay-by=··········化简ax-bx+ay-by=··········化简ax-bx+ay-by=x(a-b)+y(a-b)=(a
ax-bx+ay-by=··········化简
ax-bx+ay-by=··········化简
ax-bx+ay-by=··········化简
ax-bx+ay-by
=x(a-b)+y(a-b)
=(a-b(x+y)
ax-bx+ay-by=··········化简
ax+by-ay-bx
ax-bx-ay+by
(ax+bx+ay+by)-(ax-bx-ay-by)化简
ax+ay+bx+by(因式分解)
ax-bx-ay+by 因式分解
ax-ay-bx+by因式分解
因式分解 (ax+by+ay)^2+(bx-ay)(ax+by+ay)+(bx-ay)^2
已知,如图,在△ABC中,支线MN交AC和BC的延长线于X,Y,Z求证AX/BX·BZ/CZ·CY/AY=1
x²-y/ax-bx+ay-by=?
ax+ay-bx-by=(ax+ay)-( ）=（ ）（ ）
(AX+BY)平方+(BX-AY)平方
因式分解：(ax-by)^2+(ay+bx)^2
2ax+2ay-bx-by因式分解
（ax+by)^2+(bx-ay)^2因式分解
(ax+by)^2+(bx-ay)^2
将ay+ax-by-bx分解因式
(ax+by)^2-(ay+bx)^2